Section 4.1: Work Done by a Constant Force
Tutorial 1 Practice, page 166
1. Given: F = 275 N; !d = 0.65 m
Required: W
Analysis: Use the equation for work, W = F Δd cos θ . F and Δd are in the same direction,
so the angle between them is zero, stated as θ = 0 .
Solution: W = F!d cos"
= (275 N)(0.65 m)cos0
= (180 N # m)(1)
W = 180 N # m
Statement: The weightlifter does 1.8 ! 102 J of work on the weights.
2. Given: F = 9.4 N; Δd = 0 m
Required: W
Analysis: There is no displacement in the direction of the applied force, so no work is
done.
Statement: There is no work (0 J) done on the wall.
3. Given: F = 0.73 N; Δd = 0.080 m (note that the cue stick only does work on the ball
when it is in contact with the ball)
Required: W
Analysis: Use the equation for work, W = F Δd cos θ . F and Δd are in the same direction,
so the angle between them is zero, θ = 0 .
Solution: W = F!d cos"
= (0.73 N)(0.080 m)cos0
= (0.0584 N # m)(1)
W = 0.058 N # m
Statement: The cue stick does 0.058 J of work on the ball.
4. Given: F = 9.9 ×103 N; Δd = 4.3 m;θ = 12°
Required: W
Analysis: The work done on the car by the tow truck depends only on the component of
force in the direction of the car’s displacement. Use the equation for work,
W = F Δdcosθ . F and Δd are at an angle of 12° to each other.
Solution: W = F!d cos"
= (9.9 # 103 N)(4.3 m)cos12°
W = 4.2 # 104 N $ m
Statement: The tow truck does 4.2 ! 104 J of work on the car.
Tutorial 2 Practice, page 167
1. (a) Given: m = 56 kg; !d = 78 m; g = "9.8 m / s 2
Required: Wr, the work done by the ride on the rider
Analysis: The ride must counteract the force of gravity for it to move at a constant speed.
Fg = mg , so Fr = !mg; Wr = Fr "d cos# .
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.1-1
Solution: Wr = Fr !d cos"
= #mg!d cos"
= #(56 kg)(#9.8 m/s 2 )(78 m)cos0
= 4.3$ 104 (kg % m/s 2 )(m)
Wr = 4.3$ 104 N % m
= 4.3$ 104 J
Statement: The work done by the ride on the rider is 4.3 ! 104 J.
(b) Given: m = 56 kg; !d = 78 m; g = "9.8 m / s 2
Required: Wg, the work done by gravity on the rider
Analysis: The force of gravity on the rider is Fg = mg; Wg = Fg !dcos" .
Solution: Wg = Fg !d cos"
= mg!d cos"
= (56 kg)(#9.8 m/s 2 )(78 m)cos0
= #4.3$ 104 (kg % m/s 2 )(m)
= #4.3$ 104 N % m
Wg = #4.3$ 104 J
Statement: The work done by gravity on the rider is −4.3 ! 104 J.
2. (a) Given: F = !5.21" 103 N; #d = 355 m
Required: W
Analysis: W = F Δd cos θ
Solution: W = F!d cos"
= (#5.21$ 103 N)(355 m)cos0
= #1.85 $ 106 N % m
W = #1.85 $ 106 J
Statement: The work done by friction on the plane’s wheels is −1.85 ! 106 J.
(b) Given: F = !5.21" 103 N; W = !1.52 " 106 J
Required: Δd
W
Analysis: W = F!d cos" and W = FΔd cos θ; !d =
F cos"
W
Solution: !d =
F cos"
#1.52 $ 106 J
=
(#5.21$ 103 N)(cos0)
#1.52 $ 106 N % m
=
(#5.21$ 103 N)(cos0)
!d = 292 m
Statement: The distance travelled by the plane is 292 m.
3. Given: F = 5.9 N; ! = 150°; "d = 3.5 m
Required: W
Analysis: W = F Δd cos θ
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.1-2
Solution: W = F!d cos"
= (5.9 N)(3.5 m)(cos150°)
= #18 N $ m
W = #18 J
Statement: The work done on the skier by the snow is −18 J.
Tutorial 3 Practice, page 168
1. Given: θ = 90°
Required: W
Analysis: The gravitational pull of Earth causes the Moon to experience centripetal
acceleration during its orbit. At each moment, the Moon’s instantaneous velocity is at an
angle of 90° to the centripetal force. During a very short time interval, the very small
displacement of the Moon is also at an angle of 90° to the centripetal force. We can break
one orbit of the Moon into a series of many small displacements, each occurring during a
very short time interval. During each time interval, the centripetal force and the
displacement are perpendicular. The total work done during one orbit will equal the sum
of the work done during each small displacement. For each small segment, use the work
equation, W = F Δd cos θ , with θ = 90° .
Solution:
W = F!d cos"
= F!d cos90°
= F!d(0)
W =0 J
Statement: Summing the work done during all segments of the orbit gives a total of
W = 0 J during each revolution. The centripetal force exerted by Earth does zero work on
the Moon during the revolution.
Tutorial 4 Practice, page 169
1. Given: !d = 223 m; Fh = 122 N; " h = 37°; Ff = 72.3 N; " f = 180°
Required: Wh, Wf, WT
Analysis: W = F Δd cos θ . The total work done is the sum of the work done by the
individual forces.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.1-3
Solution: Wh = Fh !d cos"
= (122 N)(223 m)cos37°
Wh = 2.17 # 104 J (one extra digit carried)
Wf = Ff !d cos"
= (72.3 N)(223 m)cos180°
Wf = #1.61$ 104 J (one extra digit carried)
WT = Wh + Wf
= 2.17 ! 104 J + ("1.61! 104 J)
WT = 5.6 ! 103 J
Statement: The work done by the hiker is 2.2 ! 104 J. The work done by friction is
−1.6 × 104 J. The total work done is 5.6 ! 103 J.
2. Given: WT = 2.42 ! 104 J; Fh = 122 N; " h = 37°; Ff = 72.3 N; " f = 180°
Required: Δd
Analysis: WT = Wh + Wf
WT = Fh !d cos" + Ff !d cos"
WT = !d(Fh cos" + Ff cos" )
WT
!d =
Fh cos" + Ff cos"
WT
Fh cos" + Ff cos"
2.42 # 104 J
=
(122 N)(cos37°) + (72.3 N)(cos180°)
!d = 963 m
Statement: The hiker pulled the sled 963 m.
Solution: !d =
Section 4.1 Questions, page 170
1. The bottom rope does more work on the box, because it is in the same direction as the
displacement. Only the horizontal force component of the top rope does any work on the
box.
2. No. There is no work done on an object by a centripetal force, because for each small
displacement, the force is perpendicular to the direction of the displacement.
3. Given: F = 12.6 N; !d = 14.2 m; " = 21.8°
Required: W
Analysis: W = F Δd cos θ
Solution: W = F!d cos"
= (12.6 N)(14.2 m)cos 21.8°
W = 166 J
Statement: The shopper does 166 J of work on the cart.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.1-4
4. Given: F = 22.8 N; !d = 52.6 m; W = 9.53" 102 J
Required: θ
Analysis: W = F!d cos"
W
cos" =
F!d
W
Solution: cos! =
F"d
9.53# 102 J
=
(22.8 N)(52.6 m)
9.53# 102 N $ m
=
(22.8 N )(52.6 m)
= 0.7946 (one extra digit carried)
! = 37.4°
Statement: The angle between the rope and the horizontal is 37.4°.
5. (a) Given: ! = 30°; m = 24 kg; g = "9.8 m / s 2
Required: component of gravitational force along the ramp’s surface
Analysis: Draw a FBD of the situation.
Let Fg1 represent the component of gravitational force along the ramp’s surface, and Fg2
represent the component perpendicular to the ramp’s surface. Since the angle between Fg1
and Fg is 60° (from the FBD), Fg1 = Fg cos60° .
Solution: Fg1 = Fg cos60°
= mg cos60°
= (24 kg)(!9.8 m/s 2 )cos60°
Fg1 = !117.6 N (two extra digits carried)
Statement: The component of gravitational force along the ramp’s surface is 120 N
down the ramp.
(b) Analysis: The force up the ramp must exactly balance the component of gravitational
force along the ramp’s surface for the crate to move up the ramp at a constant speed.
Statement: The force required is 120 N up the ramp.
(c) Given: F = 117.6 N; Δd = 23 m
Required: W
Analysis: W = F Δd cos θ
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.1-5
Solution: W = F!d cos"
= (117.6 N)(23 m)cos0°
= 2700 J
Statement: The work done to push the crate up the ramp is 2.7 ! 103 J.
(d) Given: µk = 0.25; m = 24 kg; ! = 30°; "d = 16 m
Required: Wk; WT
Analysis: The worker has to overcome the force of friction and the force of gravity along
the ramp, so the total work done by the worker is the sum of the work done by gravity
and the work done by friction.
Solution: The work done by gravity is the force of gravity along the ramp times the
distance, which is mg cos 60°(∆d). The work done by friction is the force of friction
along the ramp times the distance, which is µkmg cos 30°(∆d). The total work done by the
worker is the sum of these two.
Ww = Wg + Wk
= (mg cos60°)(!d) + ( µk mg cos30°)(!d)
= mg!d(cos60° + µk cos30°)
= (24 kg)(9.8 m/s 2 )(16 m)(cos60° + 0.25cos30°)
Ww = 2700 J [up the ramp]
Wk = Fk !d
= µk FN !d
= µk mg(cos30°)!d
= (0.25)(24 kg)(9.8 m/s 2 )(cos30°)(16 m)
Wk = 810 J [down the ramp]
The total work done by the system is equal to the work done by the worker plus the work
done by friction.
WT = W + Wk
= 2700 J + (!810 J)
WT = 1900 J
Statement: The work done by the worker is 2.7 ! 103 J up the ramp. The work done by
kinetic friction is 8.1 × 102 J down the ramp. The total work done is 1.9 ! 103 J up the
ramp.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.1-6
6. Given: Fb = 75 N; ! b = 32°; Fg = 75 N; ! g = 22°; "d = 13 m
Required: WT
Analysis: WT = Wb + Wg ; Wb = Fb !dcos" b ; Wg = Fg !dcos" g
Solution: WT = Wb + Wg
= Fb !d cos" b + Fg !d cos" g
= (75 N)(13 m)cos32° + (75 N)(13 m)cos 22°
WT = 1700 J
Statement: The total work done by the boy and the girl together is 1.7 ! 103 J.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.1-7
Section 4.2: Kinetic Energy and the Work–Energy Theorem
Tutorial 1 Practice, page 172
1. (a) Given: vi ; vf = 2vi ; Ek i
Required: Ek f
Analysis:
1
Ek = mv 2
2
Solution:
1
Ek i = mvi2
2
1
Ek f = mvf2
2
1
= m(2vi )2
2
1
= m(4vi2 )
2
= 2mvi2
Ek f = 4Ek i
Statement: A car’s kinetic energy increases by a factor of 4 when the car’s speed
doubles.
(b) Given: vi ; vf = 3vi
Required: Ek f
Analysis:
1
Ek = mv 2
2
Solution:
1
Ek i = mvi2
2
1
Ek f = mvf2
2
1
= m(3vi )2
2
1
= m(9vi2 )
2
!1
$
= 9 # mvi2 &
"2
%
Ek f = 9Ek i
Statement: A car’s kinetic energy increases by a factor of 9 when the car’s speed triples.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.2-1
(c) Given: vi ; vf = 1.26vi ; Ek i
Required: Ek f
Analysis:
1
Ek = mv 2
2
Solution:
1
Ek i = mvi2
2
1
Ek f = mvf2
2
1
= m(1.26vi ) 2
2
1
= m(1.6vi2 )
2
⎛1
⎞
= 1.6 ⎜ mvi2 ⎟
⎝2
⎠
Ek f = 1.6 Ek i
Statement: A car’s kinetic energy increases by a factor of 1.6 when the car’s speed
increases by 26 %.
2. Given: m = 8.0 kg; v = 2.0 m/s
Required: Ek
Analysis:
1
Ek = mv 2
2
Solution:
1
Ek = mv 2
2
1
= (8.0 kg)(2.0 m/s) 2
2
= 16 kg ⋅ m 2 /s 2
Ek = 16 J
Statement: The bowling ball’s kinetic energy is 16 J.
3. Given: v = 15 km/h; Ek = 0.83 J
Required: m
Analysis:
1
Ek = mv 2
2
2 Ek = mv 2
2 Ek
v2
The speed must be converted to metres per second.
m=
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.2-2
Solution: Convert 15 km/h to metres per second.
!
km $ ! 1 h $ ! 1000 m $
= 4.17 m/s (one extra digit carried)
# 15 h & #" 3600 s &% #
" 1 km &%
"
%
2 Ek
v2
2(0.83 J)
=
(4.17 m/s) 2
m=
= 0.095
kg ⋅ m 2 /s 2
m 2 /s 2
m = 0.095 kg
Statement: The bird’s mass is 0.095 kg.
Tutorial 2 Practice, page 175
1. Given: m = 22 g = 0.022 kg; vi = 0; vf = 220 km / h
Required: W
Analysis: W = ΔEk
Solution: Convert the speed to metres per second.
km ⎞⎛ 1000 m ⎞⎛ 1 h ⎞
⎛
vf = ⎜ 220
⎟⎜
⎟⎜
⎟
h ⎠⎝ 1 km ⎠⎝ 3600 s ⎠
⎝
vf = 61.1 m/s (one extra digit carried)
W = ΔEk
= Ek f − Ek i
1 2 1 2
mvf − mvi
2
2
1
= mvf2
2
1
= (0.022 g)(61.1 m/s) 2
2
W = 41 J
Statement: The work done on the arrow by the bowstring is 41 J.
2. Given: m = 3.8 ! 104 kg; vi = 1.5 ! 104 m / s; F = 2.2 ! 105 N; "d = 2.8 ! 106 m
Required: vf
1
1
Analysis: Ekf = Eki + ΔEk ; Eki = mvi2 ; Ekf = mvf2 ; ΔEk = F Δd
2
2
=
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.2-3
Solution: Ekf = Eki + !Ek
1 2
mv + F!d
2 i
1
= (3.8 " 104 kg)(1.5 " 104 m/s)2 + (2.2 " 105 N)
2
= 4.275 " 1012 J + 6.16 " 1011 J
=
Ekf = 4.891" 1012 J (two extra digits carried)
Ekf =
1 2
mv
2 f
2
E = vf2
m kf
vf =
=
2
E
m kf
2
(4.891! 1012 J)
4
3.8 ! 10 kg
vf = 1.6 ! 104 m/s
Statement: The final speed of the probe is 1.6 ! 104 m/s.
3. Given: vi = 2.2 m / s; vf = 0; Ff = 15 N
Required: m
Analysis: Friction opposes the motion of the disc, so θ is 180°, and cos θ is –1. The work
done by friction is
W = F!d cos"
= (15 N)(12 m)cos180°
W = #180 J
Solution: The work–energy theorem tells us that the change in kinetic energy will equal
the work done, or –180 J. The final velocity is zero, so the final kinetic energy is zero.
W = Ekf ! Eki
1
W = 0 ! mvki2
2
1 2
mv = !W
2 ki
2W
m=! 2
vki
" 2(!180 J) %
= !$
# (2.2 m/s)2 '&
m = 74 kg
Statement: The skater’s mass is 74 kg.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.2-4
Section 4.2 Questions, page 176
1. Answers may vary. Sample answer:
Yes, it is possible. For example, an elephant can have a mass of up to 12 000 kg. Its slow
walking speed might be 2 m/s. Thus, its kinetic energy is
1
Ek = mv 2
2
1
= (12 000 kg)(2 m/s)2
2
= 24 000 J
A small cheetah might have a mass of 35 kg, and its top running speed is about 120 km/h,
which is about 33 m/s. Its kinetic energy is
1
Ek = mv 2
2
1
= (35 kg)(33 m/s) 2
2
= 19 000 J
Yes, it is possible that an elephant walking slowly could have greater kinetic energy than
the cheetah.
2. (a) Given: mc = 5.0 kg; mm = 0.035 kg; Ek c = 100Ek m; mouse running at a constant
speed, vm
Required: Will the cat catch up with the mouse?
Analysis: Determine the cat’s speed relative to the mouse’s speed using the fact that the
cat’s kinetic energy is 100 times the mouse’s kinetic energy.
Solution:
Ek c = 100Ek m
!1
$
1
mc vc2 = 100 # mm vm2 &
2
"2
%
mc vc2 = 100mm vm2
(5.0 kg )vc2 = 100(0.035 kg )vm2
5.0vc2 = 3.5vm2
vc2 = 0.70vm2
vc = 0.70vm
vc = 0.84vm
Statement: Since the cat’s speed is less than the mouse’s speed, the cat will never catch
up to the mouse.
(b) Analysis: The cat’s speed must be greater than the mouse’s speed for the cat to catch
up. Let the factor by which the cat’s kinetic energy is greater than the mouse’s kinetic
energy be x. Then Ekc = xEkm .
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.2-5
Solution:
Ek c = xEk m
!1
$
1
mc vc2 = x # mm vm2 &
2
"2
%
mc vc2 = xmm vm2
(5.0 kg )vc2 = x(0.035 kg )vm2
5.0vc2 = 0.035xvm2
vc2 = 0.0070xvm2
vc = 0.0070x (vm )
For the cat’s speed to be greater than the mouse’s speed, 0.0070 x > 1 .
0.0070 x > 1
0.0070 x > 1
x > 140
Statement: For the cat to catch up with the mouse, its kinetic energy must be greater than
140 times the kinetic energy of the mouse.
3. Given: m = 1.5 ! 103 kg; vi = 11 m/s; vf = 25 m/s; !d = 0.20 km
Required: W
Analysis: W = ΔEk
Solution: W = !Ek
= Ek f " Ek i
1 2 1 2
mv " mv
2 f 2 i
1
= m(vf2 " vi2 )
2
1
= (1.5 # 103 kg)((25 m/s)2 " (11 m/s)2 )
2
W = 380 000 J
Statement: The work done on the car is 3.8 ! 105 J.
4. Given: m = 9.1! 103 kg; vi = 98 km / h; vf = 27 km / h
Required: W
Analysis: W = ΔEk
Solution: Convert the speeds to metres per second.
! km $ ! 1000 m $ ! 1 h $
vi = # 98
h &% #" 1 km &% #" 3600 s &%
"
=
vi = 27.2 m/s (one extra digit carried)
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.2-6
!
km $ ! 1000 m $ ! 1 h $
vf = # 27
h &% #" 1 km &% #" 3600 s &%
"
vf = 7.5 m/s
W = !Ek
= Ek f " Ek i
1 2 1 2
mv " mv
2 f 2 i
1
= m(vf2 " vi2 )
2
1
= (9.1# 103 kg)((7.5 m/s)2 " (27.2 m/s)2 )
2
W = "3100 000 J
Statement: The work done on the truck is −3.1 ! 106 J.
5. Given: Ek1 = Ek2 ; v2 = 2.5v1
Required: m1 : m2
=
1 2
mv ;
2
Solution: Substitute the given values into the equation Ek1 = Ek2 .
Analysis: Ek =
Ek 1 = Ek 2
1
1
m1v12 = m2 v22
2
2
m1v12 = m2 (2.5v1 )2
m1 v12 = 6.25m2 v12
m1 = 6.2m2
m1
= 6.2
m2
m1 : m2 = 6.2 :1
Statement: The ratio of the slower mass to the faster mass is 6.2 : 1.
6. Given: mc = 1.2 × 103 kg; ms = 4.1 ! 103 kg; vc = 99 km/h; Ek c = Ek s
Required: vs
1
Analysis: Convert the speed to kilometres per hour. Substitute Ek = mv 2 into the
2
equation Ekc = Eks and solve for vs.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.2-7
! km $ ! 1000 m $ ! 1 h $
Solution: vc = # 99
h &% #" 1 km &% #" 3600 s &%
"
vc = 27.5 m/s (one extra digit carried)
Ek c = Ek s
1
1
mc vc2 = ms vs2
2
2
mc vc2
v =
ms
2
s
vs2 =
(1.2 ! 103 kg)(27.5 m/s)2
(4.1! 103 kg)
vs =
(1.2 ! 103 kg )(27.5 m/s)2
(4.1! 103 kg )
= 14.9 m/s (one extra digit carried)
Convert the speed back to kilometres per hour.
!
m $ ! 1 km $ ! 3600 s $
vs = # 14.9 & #
s % " 1000 m &% #" 1 h &%
"
vs = 54 km/h
Statement: The speed of the SUV is 54 km/h.
7. Given: ma = 0.020 kg; va = 250 km / h; mb = 0.14 kg; Eka = Ekb
Required: vb
1
Analysis: Ek = mv 2 ; convert speed to metres per second; substitute into Eka = Ekb
2
km 1000 m 1 h
!
!
h 1 km 3600 s
va = 69.4 m/s (one extra digit carried)
Solution: va = 250
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.2-8
Ek a = Ek b
1
1
ma va2 = mb vb2
2
2
ma va2
= vb2
mb
vb =
=
ma va2
mb
(0.020 kg )(69.4 m/s)2
0.14 kg
vb = 26.2 m/s
Convert the speed back to kilometres per hour.
!
m $ ! 1 km $ ! 3600 s $
vb = # 26.2 & #
s % " 1000 m &% #" 1 h &%
"
vb = 94 km/h
Statement: The speed of the baseball is 94 km/h.
8. Given: v = 150 km / h; m = 0.16 kg; !d = 0.25 m
Required: F
1 2
mv ; W = F!d ; W = !Ek ;
2
!Ek = Ekf " Eki . The puck has no initial velocity, so it has no initial kinetic energy, that
Analysis: Convert the speed to metres per second; Ek =
is, Eki = 0 .
km 1000 m 1 h
!
!
h 1 km 3600 s
v = 41.7 m/s (one extra digit carried)
Solution: v = 150
W = F!d
W
F=
!d
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.2-9
W = !Ek
= Ek f " Ek i
= Ek f
1 2
mv
2
1
= (0.16 kg)(41.7 m/s)2
2
W = 139 J (one extra digit carried)
=
W
!d
139 J
=
0.25 m
F = 560 N
Statement: The average force exerted on the puck by the player is 560 N.
9. Given: m = 5.31! 10"26 kg; Ek = 6.25 ! 10"21 J
Required: v
1
Analysis: Ek = mv 2 ; solve for v and substitute.
2
1
Solution: Ek = mv 2
2
2
Ek = v 2
m
F=
v=
=
2
E
m k
2
(6.25 ! 10"21 J)
"26
5.31! 10 kg
v = 485 m/s
Statement: The speed of the molecule is 485 m/s.
10. Given: Fa = 15 N;m = 3.9 kg; µk = 0.25; vi = 0.0 m / s; !d = 12 m
Required: vf
Analysis: Ff = µk FN ; FN = mg; F = Fa ! Ff ; W = "Ek ; since vi = 0.0 m / s; "Ek = Ekf ;
Ekf =
1 2
mv
2 f
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.2-10
Solution:
Ff = µk FN
= 0.25mg
= 0.25(3.9 kg)(!9.8 m/s 2 )
Ff = !9.56 N (one extra digit carried)
F = Fa + Ff
= 15 N + (!9.56 N)
F = 5.44 N (one extra digit carried)
W = F"d
= (5.44 N)(12 m)
W = 65.3 J
W = Ek
W=
1 2
mv
2 f
2
W = vf2
m
vf =
=
2
W
m
2
(65.3 J)
3.9 kg
vf = 5.8 m/s
Statement: The final speed of the block is 5.8 m/s.
11. Given: m = 5.55 ! 103 kg; v1 = 2.81 km / s or 2810 m / s; v2 = 3.24 km / s or 3240 m / s
Required: Wg
1
Analysis: Wg = !Ek ; !Ek = Ek 2 " Ek1; Ek = mv 2
2
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.2-11
Solution: Wg = !Ek
= Ek 2 " Ek 1
1 2 1 2
mv " mv
2 2 2 1
1
= m(v22 " v12 )
2
1
= (5.55 # 103 kg)((3240 m/s)2 " (2810 m/s)2 )
2
Wg = 7.22 # 109 J
=
Statement: The work done by gravity on the satellite is 7.22 ! 109 J.
12. (a) It is a quadratic function.
(b) The graph passes through the origin because when the speed is zero, the kinetic
energy is zero.
(c) Given: From the graph, when the speed, v, is 2 m/s, the kinetic energy, Ek, is 4 J.
Required: m
1
Analysis: Ek = mv 2 ; solve for m.
2
1
Solution:
Ek = mv 2
2
2
E =m
v2 k
2
m = 2 Ek
v
2
=
(4 J)
(2 m/s)2
m = 2 kg
Statement: The mass of the robot is 2 kg.
(d) Substitute the mass of the robot into the equation for kinetic energy, omitting units.
1
Ek = mv 2
2
1
= (2)v 2
2
Ek = v 2
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.2-12
Section 4.3: Gravitational Potential Energy
Tutorial 1 Practice, page 180
1. Given: m = 0.02 kg; ∆d = 8.0 m; g = 9.8 m/s2
Required: ∆Eg
Analysis: Use the gravitational potential energy equation, ΔEg = mg Δy . Let the y = 0
reference point be the ground.
Solution: !Eg = mg!y
= (0.02 kg)(9.8 m/s 2 )(8.0 m)
!Eg = 1.6 J
Statement: The change in potential energy between the branch and the ground is 1.6 J.
2. Given: ΔEg = 660 J; Δy = 2.2 m; g = 9.8 m / s2
Required: m
Analysis: Rearrange the gravitational potential energy equation, ΔEg = mg Δy , to solve
for m.
Solution: !Eg = mg!y
m=
=
!Eg
g!y
660 J
(9.8 m/s 2 )(2.2 m)
m = 31 kg
Statement: The mass of the loaded barbell is 31 kg.
3. Given: height of each book, h = 3.6 cm = 0.036 m ; number of extra books = 2
Required: W
Analysis: ΔEg = mg Δy
The 11th book is moved 10 ! 3.6 cm and the 12th book is moved 11 ! 3.6 cm.
Solution: !Eg = mg!y
= (1.6 kg)(9.8 m/s 2 )[10(0.036 m) + 11(0.036 m)]
!Eg = 12 J
Statement: The work done by the student to stack the two extra books is 12 J.
Section 4.3 Questions, page 181
1. (a) Given: m = 2.5 kg; g = 9.8 m / s 2 ; !y = 2.0 m
Required: Ek
Analysis: The kinetic energy of the wood when it hits the table is equal to the potential
energy of the wood before it falls. Ek = ΔEg = mg Δy
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.3-1
Solution: Ek = Eg
= mg!y
= (2.5 kg)(9.8 m/s 2 )(2.0 m)
Ek = 49 J
Statement: The kinetic energy of the piece of wood as it hits the table is 49 J.
(b) Given: m = 2.5 kg; Ek = 49 J
Required: v
1
2
Analysis: Ek = mv2 ; solve for v
Solution:
Ek =
1 2
mv
2
2Ek
= v2
m
v=
=
2Ek
m
2(49 J)
2.5 kg
v = 6.3 m/s
Statement: The speed of the wood as it hits the table is 6.3 m/s.
2. Given: g = 9.8 m / s2 ; m = 5.0 kg; Δy = 553 m
Required: Eg
Analysis: Eg = mg Δy
Solution: Eg = mg!y
= (5.0 kg)(9.8 m/s 2 )(553 m)
Eg = 2.7 " 104 J
Statement: The gravitational potential energy of the Canada goose is 2.7 × 104 J.
3. (a) Given: m = 175 g = 0.175 kg; !y = 1.05 m ; g = – 9.8 m/s2
Required: gravitational potential energy of the puck, Eg
Analysis: Eg = mg!y
Solution: Eg = mg!y
= (0.175 kg)(9.8 m/s 2 )(1.05 m)
Eg = 1.8 J
Statement: The gravitational potential energy of the puck is 1.8 J.
(b) Given: Eg = 1.8 J
Required: change in gravitational potential energy of puck, !Eg
Analysis: Since the gravitational potential energy of the puck when it hits the ice is equal
to 0, it is expressed as ΔEg = − Eg .
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.3-2
Solution: !Eg = " Eg
!Eg = –1.8 J
Statement: The change in gravitational potential energy of the puck is −1.8 J.
(c) Given: !Eg = –1.8 J
Required: work done by the puck, W
Analysis: Since work and energy use the same units, W is equal to the change in
gravitational potential energy of the puck.
Solution: W = !Eg
W = "1.8 J
Statement: The work done on the puck by gravity is 1.8 J.
4. The total work done is 0 J. The work done by gravity while you lift the cat is exactly
balanced by the work done by gravity while you lower the cat.
5. Given: !y = "5.4 m; !Eg = "3.1# 103 J; g = 9.8 m / s 2
Required: m
Analysis: Eg = mg Δy
Solution: At the mat, the pole vaulter’s gravitational potential energy is 0 J.
Thus, ∆Eg = −Eg.
!Eg = " Eg
!Eg = "mg!y
m="
=
!Eg
g!y
"3.1# 103 J
(9.8 m/s 2 )("5.4 m)
m = 59 kg
Statement: The pole vaulter’s mass is 59 kg.
6. Given: m = 0.46 kg; ΔEg = 155 J; g = 9.8 m / s 2
Required: ∆y
Analysis: ΔEg = mg Δy
Solution: !Eg = mg!y
!y =
=
!Eg
mg
155 J
(0.46 kg)(9.8 m/s 2 )
!y = 34 m
Statement: The maximum height of the ball above the tee is 34 m.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.3-3
7. Given: m = 59 kg; !y = 1.3 km = 1300 m; " = 14°; g = 9.8 m / s 2
Required: Eg
"y
Analysis: sin! =
; Eg = mg Δy
d
"y = d sin!
Solution: !y = d sin"
= (1300 m) sin 14°
!y = 314.498 m (four extra digits carried)
Eg = mg!y
= (59 kg)(9.8 m/s 2 )(314.498 m)
Eg = 1.8 " 105 J
Statement: The snowboarder’s gravitational potential energy is 1.8 ! 105 J.
8. (a) The work done on the first box is zero, because it doesn’t move. The second box is
lifted a height of ∆y, the third is lifted a height of 2!y , the fourth is lifted a height of
3!y , and so on until the Nth box, which is lifted a height of ( N ! 1) "y . Therefore, the
work done to raise the last box to the top of the pile is expressed as mg ( N ! 1) "y .
(b) As in Sample Problem 3 of Tutorial 1 on page 180, the gravitational potential energy
of the stack of boxes is the sum of the gravitational potential energies of the individual
boxes.
!Eg = mg[0 " !y + 1" !y + 2 " !y + 3" !y +!+ (N # 1)!y]
!Eg = mg!y[0 + 1+ 2 + 3+!+ (N # 1)]
n
The sum of an arithmetic sequence is given by the formula S n = [2a1 + (n ! 1)d] . To
2
find the sum of the sequence 0 + 1+ 2 + 3+ ... + ( N ! 1) , substitute n = N , a1 = 0 , and
d = 1.
n
S n = [2a1 + (n ! 1)d]
2
N
S N = [2(0) + (N ! 1)(1)]
2
N (N ! 1)
SN =
2
Therefore, the gravitational potential energy that is stored in the entire pile is expressed
as:
mg!yN (N " 1)
!Eg =
2
!y
!Eg = mgN (N " 1)
2
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.3-4
9. Answers may vary. Sample answer:
Given: Ec = 1.3! 108 J; 3.79 L = 1 gal; g = 9.8 m / s 2 ; 30 students in class; average mass
of each student = 70 kg
Required: ∆y
Analysis: Find the chemical potential energy, Ec 1, in 1 L of gas by dividing Ec by 3.79.
Find the total mass of the class of students. Solve the equation Eg = mg Δy for ∆y.
Ec
3.79
1.3! 108 J
=
3.79
Ec 1 = 3.43! 107 J (one extra digit carried)
Solution: Ec 1 =
There are 3.43! 107 J of chemical potential energy in 1 L of gas. Assuming that there are
30 students in the class, each with an average mass of 70 kg, this equals a total mass of
30 ! 70 kg = 2100 kg .
Therefore,
Eg = mg!y
!y =
=
Eg
mg
3.43" 107 J
(2100 kg)(9.8 m/s 2 )
!y = 1700 m
Statement: The chemical potential energy of the gas could lift the students 1700 m if it
could all be converted to gravitational potential energy.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.3-5
Section 4.5: The Law of Conservation of Energy
Mini Investigation: Various Energies of a Roller Coaster, page 185
Answers may vary. Sample answers:
A. The total energy graph is a straight line because total energy is conserved and it is
constant.
B. At any height, h, the sum of the energy values on the potential energy and kinetic
energy curves is equal to the value of the total energy at that height.
C. It is necessary to know the height at point A because it represents the potential energy
before the roller coaster starts. This is equal to the total mechanical energy. If the height
of point A were greater, the change in slopes of the potential and kinetic energy graphs
would be greater, and the total mechanical energy graph would be higher, but still
horizontal.
D. If the mass of the roller coaster car were greater, the total mechanical energy would be
greater, and the change in slopes for the graphs for potential and kinetic energy would be
greater.
Tutorial 1 Practice, page 187
1. (a) Given: m = 0.43 kg; !y = 18 m; g = 9.8 m / s 2 ; vi = 7.4 m / s
Required: vf
Analysis: The total energy at the top of the hill is equal to the total energy at the bottom
of the hill. At the top of the hill, the total energy is the gravitational potential energy,
mg Δy , plus the kinetic energy,
to the kinetic energy,
1 2
mvi . At the bottom of the hill, the total energy is equal
2
1 2
mvf , since there is no gravitational potential energy at ∆y = 0.
2
1
1
mvi2 = mvf2
2
2
2
2g!y + vi = vf2
Solution: mg!y +
vf = 2g!y + vi2
= 2(9.8 m/s 2 )(18 m) + (7.4 m/s)2
vf = 2.0 " 101 m/s
Statement: The ball’s speed at the bottom of the hill is 2.0 ! 101 m/s.
(b) Given: m = 0.43 kg; !y = 18 m; vi = 4.2 m / s; g = 9.8 m / s 2
Required: vf
Analysis: The total energy when the ball is kicked up the hill is equal to the total energy
when the ball reaches the bottom of the hill. The energy at any time when the ball is on
1
its way up or down the hill does not matter. When the ball is kicked, ET = mg!y + mvi2 .
2
1
At the bottom of the hill, ET = mvf2 .
2
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.5-1
1
1
mvi2 = mvf2
2
2
2
2g!y + vi = vf2
Solution: mg!y +
vf = 2g!y + vi2
= 2(9.8 m/s 2 )(18 m) + (4.2 m/s)2
vf = 19 m/s
Statement: The ball’s speed as it reaches the bottom of the hill is 19 m/s.
2. (a) Given: m = 0.057 kg; !y = 1.8 m; g = 9.8 m / s 2 ; vf = 0 m / s
Required: vi
Analysis: Let the player’s hand be the y = 0 reference point. The total energy when the
1 2
mv . The total energy at the highest point of the
2 i
1
ball is all gravitational potential energy, mg Δy . Thus, mvi2 = mg!y .
2
1
Solution: mvi2 = mg!y
2
vi2 = 2g!y
ball is released is all kinetic energy,
vi = 2g!y
= 2(9.8 m/s 2 )(1.8 m)
vi = 5.9 m/s
Statement: The speed of the ball as it leaves the player’s hand is 5.9 m/s.
(b) Given: m = 0.057 kg; vi 2 = vi ; g = 9.8 m / s 2
Required: Δy2 : Δy1
Analysis:
1 2
mv = mg!y
2 i
Solution:
1
mvi21 = mg!y1
2
1 2
v = g!y1
2 i1
v2
!y1 = i 1
2g
1
mvi22 = mg!y2
2
1 2
v = g!y2
2 i2
v2
!y2 = i 2
2g
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.5-2
vi22
!y2 2g
=
!y1 vi21
2g
=
=
vi22
2g
"
2g
vi21
vi22
vi21
#1 &
%$ 4 vi 1 ('
=
vi21
2
1 2
vi 1
16
=
vi21
!y2 1
=
!y1 16
Statement: The ratio of the maximum rise of the ball after leaving the player’s hand to
the maximum rise in (a) is 1:16.
Tutorial 2 Practice, page 190
1. (a) Given: v = 1.4 m / s; Δy = 5.0 m; m = 65 kg; g = 9.8 m / s 2
Required: P
Analysis: The work done to get to the top of the ladder is equal to the gravitational
potential energy at the top of the ladder, W = mg Δy . The time, t, taken to get to the top of
the ladder is the distance, ∆y, divided by the speed, v. P =
W
t
!y
v
5.0 m
=
1.4 m/s
t = 3.57 s (one extra digit carried)
Solution: t =
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.5-3
W
t
mg!y
=
t
(65 kg)(9.8 m/s 2 )(5.0 m)
=
3.57 s
P = 890 W
Statement: The firefighter’s power output while climbing the ladder is 890 W.
(b) From (a), it takes the firefighter 3.6 s to climb the ladder.
2. Given: vf 2 = 2vf 1; t2 = t1; vi = 0 m / s; m2 = m1
Required: ratio of power needed, P2 : P1
Analysis: We are told that the Grand Prix car accelerates to twice the speed of the car in
Sample 1, which can be expressed as vf 2 = 2vf 1 . We are also told that the Grand Prix car
accelerates to this speed in the same amount of time as the car in Sample 1, which is
stated as 7.7 s. We will assume that the two cars are equal in mass, at 1.1! 103 kg .
P=
Solution: P1 =
mvf21
2t
P2 =
mvf22
2t
mvf22
P2
2t
=
P1 mvf21
2t
=
2
f2
2
f1
v
v
(2vf 1 )2
=
vf21
=
4 vf21
vf21
P2 4
=
P1 1
Statement: The ratio of the power needed by the Grand Prix car to the power needed by
the car in Sample Problem 1 is 4:1.
3. Given: Δd = 190 m; t = 4 min 50 s = 290 s; m = 62 kg; g = 9.8 m / s 2
Required: P
W
Analysis: P = ; W = mg!y
t
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.5-4
W
t
mg!y
=
t
(62 kg)(9.8 m/s 2 )(190 m)
=
290 s
P = 0.40 kW
Statement: The racer’s average power output during the race is 0.40 kW.
Solution: P =
Section 4.5 Questions, page 191
1. (a) Given: vi = 11 m / s; g = 9.8 m / s 2
Required: maximum height that the ball will reach, ∆y
Analysis: The kinetic energy when the child tosses the ball is equal to the gravitational
1
potential energy at the ball’s maximum height, expressed as mvi2 = mg!y .
2
1
Solution: mvi2 = mg!y
2
1 2
v = g!y
2 i
v2
!y = i
2g
=
(11 m/s)2
2(9.8 m/s 2 )
!y = 6.2 m
Statement: The maximum height that the ball will reach is 6.2 m.
(b) As the ball leaves the child’s hand, the gravitational potential energy is zero.
It increases quadratically to its maximum when the ball reaches its maximum height.
It decreases quadratically to zero as the ball returns to the level of the child’s hand.
(c) The graph has this shape because, as the ball leaves the child’s hand, the kinetic
energy is at its maximum. It decreases quadratically to zero when the ball reaches its
maximum height. It increases quadratically to its maximum as the ball returns to the level
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.5-5
of the child’s hand. The total energy is conserved, so it is a constant horizontal line equal
to the maximum kinetic energy or potential energy.
2. Answers may vary. Sample answers:
(a) The kinetic energy is the greatest just before the apple hits the ground.
(b) The gravitational potential energy is the greatest as the apple leaves the branch.
3. (a) The law of conservation of energy states that energy can neither be created nor
destroyed in an isolated system; it can only change form. Assuming the puck and surface
form an isolated system, the energy of the hockey puck is conserved. The kinetic energy
of the puck is transformed to thermal energy by friction.
(b) The initial kinetic energy is transformed to thermal energy by friction as the puck
slows down to a stop.
4. (a) Given: m = 110 kg; !y = 210 m; g = 9.8 m / s 2
Required: W
Analysis: The work done by gravity is equal to the gravitational potential energy at the
top of the hill, expressed as Eg = mg Δy .
Solution: W = Eg
= mg!y
= (110 kg)(9.8 m/s 2 )(210 m)
W = 2.3" 105 J
Statement: The work done by gravity on the skier is 2.3 ! 105 J.
(b) Given: m = 110 kg; !y = 210 m; g = 9.8 m / s 2 ; vi = 0 m / s
Required: vf
Analysis: Because the skier has no initial velocity, the total energy at the top of the hill is
all potential energy. The total energy at the bottom of the hill is all kinetic energy. The
total energy at the top of the hill is equal to the total energy at the bottom of the hill, or
1
mg!y = mvf2 . Solve for vf and substitute.
2
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.5-6
1
mv 2
2 f
2g!y = vf2
Solution: mg!y =
vf = 2g!y
= 2(9.8 m/s 2 )(210 m)
vf = 64 m/s
Statement: The skier’s speed when he reaches the bottom of the hill is 64 m/s.
5. Given: m = 62 kg; vi = 8.1 m / s; g = 9.8 m / s 2 ; !y = 3.7 m
Required: vf
Analysis: The total energy as the snowboarder leaves the ledge is the sum of her
gravitational potential energy, mg Δy , and her kinetic energy,
when she lands is all kinetic energy,
mg!y +
1 2
mvi . The total energy
2
1 2
mvf , if we take Δy = 0 at the landing point. Thus,
2
1 2 1 2
mv = mv . Solve for vf and substitute the given values.
2 i 2 f
1
1
mvi2 = mvf2
2
2
2
2g!y + vi = vf2
Solution: mg!y +
vf = 2g!y + vi2
= 2(9.8 m/s 2 )(3.7 m) + (8.1 m/s)2
vf = 12 m/s
Statement: The snowboarder’s speed at the moment she hits the ground is 12 m/s.
6. Given: !y = 3.5 m; " = 40°; g = 9.8 m / s 2
Required: the speed in the y-direction
1
Analysis: mg!y = mv 2
2
The energy equations give the speed in the direction along the jump, so we need to use
v
components to solve for the vertical velocity: v y =
.
sin !
1
Solution: mg!y = mv 2
2
2g!y = v 2
v = 2g!y
= 2(9.8 m/s 2 )(3.5 m)
v = 8.28 m/s (one extra digit carried)
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.5-7
v
sin !
8.28 m/s
=
sin 40°
v y = 13 m/s
vy =
Statement: The dolphin’s minimum speed is 13 m/s.
7. (a) Yes, the mechanical energy of the roller coaster is conserved because there is no
friction.
(b) Given: m = 640 kg; vi = 0;!yA = 30.0 m; g = 9.8 m / s 2
Required: ET, the total mechanical energy
Analysis: Because the car starts from rest, its total mechanical energy is equal to its
potential energy at point A: ET = mg ΔyA .
Solution: ET = mg!yA
= (640 kg)(9.8 m/s 2 )(30.0 m)
ET = 1.9 " 105 J
Statement: The total mechanical energy at point A is 1.9 ! 105 J.
(c) The total mechanical energy is conserved, so it is the same at point B as it is at point
A: 1.9 ! 105 J.
(d) Given: m = 640 kg;!yB = 15.0 m; g = 9.8 m / s 2 ; !yA = 30.0 m
Required: vB ; vC
Analysis for vB: The total mechanical energy at point B is the sum of the kinetic
1
2
energy, mvB2 , and the potential energy, mg!yB . The total mechanical energy is equal to
the potential energy at point A: mg!yA . Therefore,
Solution for vB:
1 2
mv + mg!yB = mg!yA .
2 B
1
mvB2 + mg!yB = mg!yA
2
vB2 + 2g!yB = 2g!yA
vB2 = 2g!yA " 2g!yB
vB = 2g(!yA " !yB )
= 2(9.8 m/s 2 )(30.0 m " 15 m)
vB = 17 m/s
Statement for vB: The speed of the car when it reaches point B is 17 m/s.
1 2
mv .
2 C
The total mechanical energy is equal to the potential energy at point A: mg∆yA.
1
Thus, mvC2 = mg!yA .
2
Analysis for vC: The total mechanical energy at point C is all kinetic energy,
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.5-8
Solution for vC:
1
mv 2 = mg!yA
2 C
vC2 = 2g!yA
vC = 2g(!yA )
= 2(9.8 m/s 2 )(30.0 m)
vC = 24 m/s
Statement for vC: The speed of the car when it reaches point C is 24 m/s.
(e) Given: !yA = 30.0 m; !yB = 15.0 m; g = 9.8 m / s 2
Required: vB ; vC
Analysis: The total energy at A is equal to the total energy at B and at C. At A, the total
energy consists of kinetic energy and gravitational potential energy. At B, the total
energy also consists of kinetic energy and gravitational potential energy. At C, the total
1
energy consists of kinetic energy only: Ek = mv 2 ; Eg = mg!y
2
Solution:
ET B = ET A
mg!yB +
1
1
mvB2 = mg!yA + mvA2
2
2
1 2
1
vB = g!yA + vA2 " g!yB
2
2
2
vB = 2g!yA + vA2 " 2g!yB
vB = 2g!yA + vA2 " 2g!yB
= 2(9.8 m/s 2 )(30.0 m) + (12 m/s)2 " 2(9.8 m/s 2 )(15.0 m)
vB = 21 m/s
ET C = ET A
1
1
mvC2 = mg!yA + mvA2
2
2
1 2
1
vC = g!yA + vA2
2
2
2
vC = 2g!yA + vA2
vC = 2g!yA + vA2
= 2(9.8 m/s 2 )(30.0 m) + (12 m/s)2
vC = 27 m/s
Statement: The speed at B is 21 m/s, and the speed at C is 27 m/s.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.5-9
8. Given: m = 52 kg; t = 24 s; !y = 18 m; g = 9.8 m / s 2
Required: P
W
Analysis: P = ; W = mg!y
t
W
Solution: P =
t
mg!y
=
t
(52 kg)(9.8 m/s 2 )(18 m)
=
24 s
P = 380 W
Statement: The power the woman exerts is 3.8 ! 102 W.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.5-10
Section 4.6: Elastic Potential Energy and Simple
Harmonic Motion
Mini Investigation: Spring Force, page 193
Answers may vary. Sample answers:
A. The relationship between Fg and ∆x is linear.
B. The slope of the best fit line of my graph is 50. This line represents the relationship
between Fg and ∆x, where the slope is the spring constant.
C. For the equipment used in this investigation, where k is the slope of the line of best fit,
the equation is Fg = 50Δx .
Tutorial 1 Practice, page 195
1. (a) Given: m = 0.65 kg; !x = 0.44 m; g = 9.8 m / s 2
Required: k
Analysis: The force of gravity on the mass points down. The restorative spring force on
the mass points up because the spring is stretched down. To calculate the total force,
subtract !the magnitudes:
!
!
Fg = mg [down] = !mg [up] ; Fx = !k"x = k"x [up]
!
!
F
= 0 according to Newton’s second law.
Since the mass is not
accelerating,
!
Solution:
!F = 0
k"x # mg = 0
mg
k=
"x
(0.65 kg)(9.8 m/s 2 )
=
0.44 m
k = 14.5 N/m (one extra digit carried)
Statement: The spring constant is 14 N/m.
(b) Given: k = 14.5 N / m; !x = 0.74 m; g = 9.8 m / s 2
Required: m
Analysis: Use the equation k!x " mg = 0 from (a).
Solution: k!x " mg = 0
k!x = mg
k!x
m=
g
(14.5 N/m)(0.74 m)
9.8 m/s 2
m = 1.1 kg
Statement: The new mass is 1.1 kg.
2. Given: m = 5.3 kg; k = 720 N / m; !x = 0.36 m
! !
Required: Fnet ; a
Analysis: The free-body diagram for the mass is shown.
=
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.6-1
The force of gravity on the mass points down. The spring force on the mass points down
because !the spring is compressed
upward:
!
!
Fg = mg [down] , Fx = !k"x = k"x [down]
!
! !
Fnet = Fg + Fx
!
!
Use the equation Fnet = ma to find the acceleration.
!
! !
Solution: Fnet = Fg + Fx
= mg [down] + k!x [down]
= (5.3 kg)(9.8 m/s 2 ) [down] + (720 N/m)(0.36 m) [down]
!
Fnet = 311 N (one extra digit carried)
!
!
Fnet = ma
!
! Fnet
a=
m
311 N [down]
=
5.3 kg
!
a = 59 m/s 2 [down]
Statement: The force on the mass is 310 N [down], and the acceleration is
59 m/s2 [down].
Tutorial 2 Practice, page 196
1. Given: mb = 2m; k = 2.29 ! 103 N / m; g = 9.8 m / s 2
Required: ratio of Eeb : Ee
1
2
Analysis: E = k (Δx)2
To find Eeb , we need Δxb . Use the same equation as in Sample Problem 1,
k!x [up] " mg [down] = 0 .
Solution: k!xb [up] " mb g [down] = 0
k!xb = mb g
mb g
k
2mg
!xb =
k
!xb =
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.6-2
Ee b =
1
k(!xb )2
2
1 " 2mg %
= k$
2 # k '&
Ee =
2
" 2
%
1 $ 4 m2 g 2 '
= k
2 $ k2 '
#
&
2m2 g 2
Ee b =
k
1
k(!x)2
2
1 " mg %
= k$
2 # k '&
=
Ee =
2
1 " m2 g 2 %
k
2 $# k 2 '&
m2 g 2
2k
2 m2 g 2
Ee b
k
=
2
Ee
m g2
2k
Ee b 4
=
Ee 1
Statement: The ratio of the elastic potential energy, Eeb : Ee is 4:1.
!
2. Given: Fx = 220 N; !x = 0.14 m
Required: Ee
!
1
!
Analysis: Fx = !k"x = k"x ; Ee = k(!x)2
2
!
Solution: Fx = k!x
!
Fx
k=
!x
220 N
=
0.14 m
k = 1570 N/m (one extra digit carried)
1
k(!x)2
2
1
= (1570 N/m)(0.14 m)2
2
Ee = 15 J
Statement: The elastic potential energy of the toy is 15 J.
Ee =
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.6-3
Tutorial 3 Practice, page 199
1. Given: m = 105 kg; k = 8.1! 103 N / m
Required: f, T
Analysis: Use the equations for simple harmonic motion period and frequency:
T = 2π
1
m
and f =
k
T
Solution: T = 2!
m
k
105 kg
7.6 " 103 N/m
T = 0.74 s
= 2!
f =
1
T
1
0.74 s
f = 1.4 Hz
Statement: The period of the vibrations is 0.74 s, and the frequency is 1.4 Hz.
2. The frequency of oscillations will change if passengers are added to the car because
when the mass increases, the period increases. This happens because mass is in the
numerator of the equation for the period. If the period increases, the frequency decreases,
because frequency is the reciprocal of period.
=
Section 4.6 Questions, page 200
1. Spring A!is more difficult to stretch because it has a greater spring constant.
2. Given: F = 5 N; !x = 10 mm = 0.01 m
Required: k
!
Analysis: The spring force opposes the applied force, so Fx = !5 N . Rearrange the
!
formula Fx = !k"x to solve for k.
!
Solution: Fx = !k"x
!
Fx
k=!
"x
(!5 N)
=!
0.01 m
k = 500 N/m
Statement: The spring constant is 500 N/m.
3. The elastic potential energy stored in a spring is the same whether it is stretched by
1.5 cm or compressed by 1.5 cm. The spring constant is exactly the same whether the
spring is stretched or compressed, so the elastic potential energy must also be the same.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.6-4
4. (a) Given: k = 5.5 ! 103 N / m; "x = 2.0 cm = 0.020 m
Required: Ee
1
Analysis: Ee = k(!x)2
2
1
Solution: Ee = k(!x)2
2
1
= (5.5 " 103 N/m)(0.020 m)2
2
Ee = 1.1 J
Statement: The elastic potential energy of the spring when it stretches 2.0 cm is 1.1 J.
(b) Given: k = 5.5 ! 103 N / m; "x = #3.0 cm = #0.030 m
Required: Ee
1
Analysis: Ee = k(!x)2
2
1
Solution: Ee = k(!x)2
2
1
= (5.5 " 103 N/m)(#0.030 m)2
2
Ee = 2.5 J
Statement: The elastic potential energy of the spring when it compresses 3.0 cm is 2.5 J.
5. (a) Given: m = 0.63 kg; k = 65 N / m; g = 9.8 m / s 2
Required: ∆x
Analysis: The force of gravity on the mass points down. The restorative force on the
mass
! points up since the spring!is compressed.
!
Fg = mg [down] = !mg [up] ; Fx = !k"x = k"x [up]
!
Solution: Since the mass is at rest, !F = 0 .
!
!F = 0
k"x # mg = 0
mg
"x =
k
(0.63 kg)(9.8 m/s 2 )
=
65 N/m
"x = 0.095 m
Statement: The spring is compressed 0.095 m from its equilibrium position.
(b) Given: m = 0.63 kg; k = 65 N / m; g = 9.8 m / s 2
!
Required: a
Analysis: The force of gravity on the mass points down. The spring force on the mass
points
up because the
being compressed downward.
!
! mass is
!
Fg = mg [down] ; Fx = !k"x = k"x [up]
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.6-5
!
! ! !
!
Fnet = Fg + Fx ; Fnet = ma
Solution:!
! !
Fnet = Fg + Fx
= mg [down] + k!x [up]
= (0.63 kg)(9.8 m/s 2 ) [down] + (65 N/m)(0.041 cm) [up]
!
Fnet = 3.5 N [down]
!
!
Fnet = ma
!
! Fnet
a=
m
3.5 N [down]
=
0.63 kg
!
a = 5.6 m/s 2 [down]
Statement: The acceleration of the mass after it falls 4.1 cm is 5.6 m/s2 [down].
6. Given: m = 5.2 kg; T = 1.2 s
Required: k
m
k
Analysis: T = 2!
Solution:
m
k
T = 2!
T
=
2!
m
k
2
" T %
m
$# 2! '& = k
k=
=
m
" T %
$# 2! '&
2
5.2 kg
" 1.2 s %
$# 2! '&
2
k = 140 N/m
Statement: The spring constant is 140 N/m.
7. Given: k = 1.5 ! 103 N / m; Ee = 80.0 J
Required: ∆x
1
Analysis: Ee = k(!x)2
2
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.6-6
Ee =
Solution:
1
k(!x)2
2
2Ee
= (!x)2
k
!x =
2Ee
k
2(80.0 J)
1.5 " 103 N/m
!x = 0.33 m
Statement: The spring should be stretched 0.33 m to store 80.0 J of energy.
8. Given: !x = 15 mm = 0.015 m; k = 400.0 N / m
Required: W
Analysis: The work done is equal to the elastic potential energy:
1
W = Ee ; Ee = k(!x)2
2
1
Solution: Ee = k(!x)2
2
1
= (400.0 N/m)(0.015 m)2
2
Ee = 0.045 J
Statement: The work done by the spring force acting on the spring is 4.5 ! 10−2 J.
9. Given: Ee = 7.50 J; m = 0.20 kg; k = 240 N / m
Required: f; ∆x
1
m
1
Analysis: T = 2!
; f = ; Ee = k (Δx)2
2
T
k
=
Solution: T = 2!
m
k
0.20 kg
240 N/m
T = 0.181 s (one extra digit carried)
= 2!
f =
1
T
1
0.181 s
f = 5.5 Hz
=
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.6-7
Ee =
1
k(!x)2
2
2Ee
= (!x)2
k
!x =
2Ee
k
2(7.50 J)
240 N/m
!x = 0.25 m
Statement: The frequency of oscillation is 5.5 Hz and the amplitude of oscillation
is 0.25 m.
10. Given: m = 5.5 ! 102 kg ; six cycles in 4.4 s
Required: k
m
Analysis: Divide 4.4 s by 6 to get T. Use the formula for the period, T = 2!
, to
k
calculate k.
4.4 s
Solution: T =
6
=
T = 2!
m
k
T
m
=
2!
k
2
T
m
=
2
k
4!
4! 2 m
k=
T2
4! 2 (5.5 " 102 kg)
=
2
# 4.4 s &
%$ 6 ('
k = 4.0 " 104 N/m
Statement: The spring constant of either spring is 4.0 × 104 N/m.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.6-8
11. Answers may vary. Sample answer: Pyon pyon shoes strap onto the outside of your
regular shoes. They are made of two curved springy pieces of material joined together in
a shape similar to that of a football. When you jump, the springy material increases the
height you can attain. When your mass presses down on the springs, the springs press
back up, causing you to jump up in the air much higher than normal.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.6-9
Section 4.7: Springs and Conservation of Energy
Tutorial 1 Practice, page 205
1. If the ramp is not frictionless, some of the kinetic energy of the block will be
transformed to thermal energy by friction as the block slides down the ramp. Thus, the
block will have less kinetic energy to compress the spring, and the amount of
compression will be less.
2. Given: m = 3.5 kg; !y = 2.7 m; !x = 26 cm = 0.26 m
Required: k
1
Analysis: Eg = mg∆y; Ee = k(!x)2
2
Since energy is conserved, the change in potential energy of the mass must equal the
change in elastic potential energy when the spring is compressed.
Solution: If we choose the bottom of the ramp to be the y = 0 reference point, the mass
will have no gravitational potential energy at the bottom of the ramp. The initial
gravitational potential energy has transformed into kinetic energy. When the spring is
fully compressed, the kinetic energy has transformed into elastic potential energy.
Therefore, the spring’s initial gravitational potential energy must equal its final elastic
potential energy.
Ee = Eg
1
k(!x)2 = mg!y
2
2mg!y
k=
(!x)2
=
2(3.5 kg)(9.8 m/s 2 )(2.7 m)
(0.26 m)2
k = 2700 N/m
Statement: The spring constant is 2.7 ! 103 N/m.
3. Given: m = 43 kg; k = 3.7 kN / m = 3700 N / m; !x = 37 cm = 0.37 m
Required: Δy
1
Analysis: !Eg = mg!y ; Ee = k(!x)2
2
Solution: Choose the lowest point of the bounce as the y = 0 reference point. At the
maximum height, ∆y, all of the elastic potential energy has converted to gravitational
potential energy.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.7-1
!Eg = Ee
1
k(!x)2
2
k(!x)2
!y =
2mg
mg!y =
=
=
(3700 N/m)(0.37 m)2
2(43 kg)(9.8 m/s 2 )
#
m&
3700
kg
"
(0.37 m)
%
s 2 ('
$
2(43 kg )(9.8 m/s 2 )
!y = 0.60 m
Statement: The maximum height he reaches on the following jump is 0.60 m above the
compressed point.
4. Given: m = 0.35 kg; h = 2.6 m; !x = 0.14
Required: k
1
Analysis: Eg = mg!y ; Ee = k(!x)2
2
The initial gravitational potential energy of the branch is equal to the final elastic
potential energy of the trampoline at its lowest point.
Solution: Choose the lowest point of the trampoline as the y = 0 reference point. At this
point, all of the gravitational potential energy has transformed to elastic potential energy.
Since the lowest point represents y = 0 , the change in y is the height above the
trampoline surface, 2.6 m, plus the maximum compression of the trampoline, 0.14 m.
Therefore, Δy = 2.6 m + 0.14 m = 2.74 m .
Ee = Eg
1
k(!x)2 = mg!y
2
2mg!y
k=
(!x)2
=
2(0.35 kg)(9.8 m/s 2 )(2.74 m)
(0.14 m)2
k = 960 N/m
Statement: The spring constant is 960 N/m.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.7-2
5. (a) Given: m = 4.0 kg; ∆y = 0.308 m
Required: v
1
Analysis: Eg = mg!y ; Ek = mv 2
2
When the mass is doubled at the top of the ramp, it is at rest, so it has only gravitational
potential energy. At the bottom of the ramp, on the horizontal segment, all of the
gravitational potential energy has transformed to kinetic energy.
Solution: Let the bottom of the ramp be the y = 0 reference point. Therefore, the
gravitational potential energy at the top of the ramp is equal to the kinetic energy along
the horizontal part of the ramp.
Ek = Eg
1
mv 2 = mg!y
2
v = 2g!y
= 2(9.8 m/s 2 )(0.308 m)
v = 2.5 m/s
Statement: The speed of the block as it returns along the horizontal surface is 2.5 m/s.
(b) No, the block does not have the same kinetic energy as before along the horizontal
surface. The kinetic energy is equal to the gravitational potential energy, which is greater
than before.
(c) Given: md = 2m
Required: Δxd
Analysis: The elastic potential energy of the spring at its greatest compression is equal to
the gravitational potential energy at the top of the ramp.
Solution: Block in Sample Problem 3: Block with Mass Doubled
1
mg!y = k(!x)2
2
!x =
2mg!y
k
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.7-3
1
k(!xd )2
2
1
2mg!y = k(!xd )2
2
md g!y =
!xd =
=
4mg!y
k
2(2mg!y)
k
= 2
2mg!y
k
!xd = 2(!x)
Since the gravitational potential energy is doubled, the compression increases, but it is
not doubled. It increases by a factor of 2. So the new value of ∆x is
0.31 m.
Statement: The new value of ∆x is 31 cm.
(d) Given: µk = 0.15; !y = 0.308 m;!d = 0.62 m; k = 250 N / m
Required: ∆x
!
!
Analysis: Ff = µ FN = µ mg ; Wf = Ff !d ; Ek = Eg ! Wf ; Ee = Ek
2(0.22 m) or
Solution: Let the bottom of the ramp be the y = 0 reference point.
!
Ff = µ mg
= 0.15(4.0 kg)(9.8 m/s 2 )
!
Ff = 5.88 N (one extra digit carried)
Wf = Ff !d
= (5.88 N)(0.62 m)
Wf = 3.65 J (one extra digit carried)
Ek = Eg ! Wf
= mg"y ! Wf
= (4.0 kg)(9.8 m/s 2 )(0.308 m) ! 3.65 J
Ek = 8.42 J (one extra digit carried)
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.7-4
Ee = Ek
1
k(!x)2 = 8.42 J
2
!x =
=
2(8.42 J)
k
2(8.42 N " m)
250 N/m
!x = 0.26 m
Statement: The new value of the compression is 0.26 m.
Research This: Perpetual Motion Machines, page 206
Answers may vary. Sample answers:
A. I chose a metronome. A spring is wound tight, and as it unwinds, the elastic potential
energy is converted to kinetic energy, forcing the metronome wand to swing. As the
wand swings to its highest point, kinetic energy transforms to potential energy. As the
wand swings down again, the gravitational potential energy converts back to kinetic
energy. The spring contributes kinetic energy at the bottom of each swing, until the
spring is fully unwound. Eventually the metronome stops due to air resistance.
B. The design of the machine has been improved over time by creating quartz
metronomes, electronic metronomes, computer metronomes, and even metronome apps
for smart phones.
C. The improvements have been the results of all three developments: new materials;
new technology; and new scientific discoveries.
Section 4.7 Questions, page 208
1. The total mechanical energy of the system increases. Energy has been added by the
person outside the system of the mass and spring.
2. Given: k = 520 N / m; m = 4.5 kg; !x = 0.35 m
Required: v
Analysis: When the spring is compressed, but the mass is at rest, the mass has only
elastic potential energy. As the spring is released, the elastic potential energy transforms
to kinetic energy. When the mass is no longer touching the spring, all of the energy is
kinetic.
1
1
Ee = k(!x)2 ; Ek = mv 2
2
2
Solution: The kinetic energy when the mass leaves the spring is equal to the elastic
potential energy when the spring is at its maximum compression.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.7-5
Ek = Ee
1 2 1
mv = k(!x)2
2
2
v=
=
k(!x)2
m
(520 N/m)(0.35 m)2
4.5 kg
v = 3.8 m/s
Statement: The speed of the mass when it leaves the spring is
3.8 m/s [away from the spring].
3. (a) Given: k = 5.2 ×102 N / m; Δx = 5.2 cm = 0.052 m
Required: Ee
1
Analysis: Ee = k(!x)2
2
1
Solution: Ee = k(!x)2
2
1
= (5.2 " 102 N/m)(0.052 m)2
2
Ee = 0.703 J (one extra digit carried)
Statement: The elastic potential energy of the compressed spring is 0.70 J.
(b) Given: Ee = 0.703 J; m = 8.4 g = 0.0084 kg; ∆x = 5.2 cm = 0.052 m
Required: v
1
Analysis: Eg = mg∆y; Ek = mv 2
2
All of the elastic potential energy of the compressed spring has transformed to
gravitational potential energy and kinetic energy as the pilot ejects.
Solution: The kinetic energy of the pilot as it ejects and the additional gravitational
potential energy is equal to the elastic potential energy of the compressed spring.
Eg + Ek = Ee
1
mg ! x + mv 2 = Ee
2
v=
2Ee ! 2mg ! x
m
2(0.703 J) ! 2(0.084 kg)(9.8 m/s 2 )(0.052 m)
=
0.0084 kg
=
2(0.660 194 kg " m 2 /s 2 )
0.0084 kg
v = 13 m/s
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.7-6
Statement: The speed of the pilot as it ejects upward from the airplane is 13 m/s above
the launch point of the compressed spring.
(c) Given: m = 8.4 g = 0.0084 kg; k = 5.2 ! 102 N / m; "x = 5.2 cm = 0.052 m
Required: Δy
Analysis: Let the point where the pilot ejects be the y = 0 reference point. All of the
elastic potential energy of the compressed spring has transformed to gravitational
potential energy at the pilot’s maximum height.
1
Ee = k (Δx)2 ; Eg = mg!y
2
Solution: The elastic potential energy of the compressed spring is equal to the
gravitational potential energy at the pilot’s maximum height.
Eg = Ee
1
k(!x)2
2
k(!x)2
!y =
2mg
mg!y =
=
$
m'
2
5.2
"
10
kg
#
(0.052 m) 2
&
2 )
s (
%
2(0.0084 kg )(9.8 m/s 2 )
!y = 8.5 m
Statement: The maximum height that the pilot will reach is 8.5 m above the launch point
of the compressed spring.
4. Given: k = 1.2 ! 102 N / m; m = 82 g = 0.082 kg; "y = 3.4 cm = 0.034 m
Required: Δx
Analysis: The elastic potential energy of the spring transforms to kinetic energy and then
to gravitational potential energy as it comes to rest at the top of the ramp.
1
Ee = k(!x)2 ; Eg = mg!y
2
Solution: Let the bottom of the ramp represent the y = 0 reference point. The elastic
potential energy of the spring is equal to the gravitational potential energy at the top of
the ramp.
Ee = Eg
1
k(!x)2 = mg!y
2
!x =
2mg!y
k
2(0.082 kg)(9.8 m/s 2 )(0.034 m)
1.2 " 102 N/m
!x = 0.021 m
Statement: The distance of the spring’s compression is 0.021 m.
=
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.7-7
5. Given: m = 75 kg; k = 6.5 N / m
Required: v
Analysis: Let the y = 0 reference point be 19 m below the platform. Since the unstretched
bungee cord is 11 m long, and the cord is stretched 19 m below the
platform, Δx = 19 m − 11 m = 8 m . The gravitational potential energy is transformed to
elastic potential energy and kinetic energy at this point.
1
1
Eg = mg Δy ; Ek = mv 2 ; Ee = k(!x)2
2
2
Solution: The gravitational potential energy at the platform is equal to the sum of the
kinetic energy and the elastic potential energy 19 m below the platform.
Eg = Ek + Ee
1 2 1
mv + k(!x)2
2
2
1 2
1
mv = mg!y " k(!x)2
2
2
mg!y =
v=
=
&
2#
1
mg!y " k(!x)2 (
%
m$
2
'
,
2 )
1
(75 kg)(9.8 m/s 2 )(19 m) " (65.5 N/m)(8.0 m)2 .
+
(75 kg) *
2
-
v = 18 m/s
Statement: The speed of the bungee jumper at 19 m below the bridge is 18 m/s.
6. Given: k = 5.0 N / m; m = 0.25 kg; !x = 14 cm = 0.14 m
Required: hmax ; vmax ; amax
Analysis: Let the y = 0 reference point be the rest position of the spring. In simple
harmonic motion, the maximum height is the opposite of the lowest point.
The maximum velocity occurs as the box passes through the rest position. At this point,
1
1
there is only kinetic energy. Thus, k(!x)2 = mv 2 .
2
2
The maximum acceleration occurs at the maximum height. At this point, the spring force
is equal to the applied force, so k!x = ma .
Solution: The maximum height is 14 cm [above rest position].
For the maximum velocity,
1
1
k(!x)2 = mv 2
2
2
v=
k(!x)2
m
(5.0 N/m)(0.14 m)2
=
0.25 kg
v = 0.63 m/s
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.7-8
For the maximum acceleration,
k!x = ma
k!x
m
(5.0 N/m)(0.14 m)
=
0.25 kg
a=
a = 2.8 m/s 2
Statement: The maximum height is 14 cm [above rest position]. The maximum speed is
0.63 m/s. The maximum acceleration is 2.8 m/s2 [toward rest position].
7. Given: m = 0.22 kg; k = 280 N / m; !x = 11 cm = 0.11 m
Required: h
Analysis: Let the y = 0 reference point be the fully compressed position of the spring.
When the block is dropped, all the energy is in the form of gravitational potential energy.
At the full compression of the spring, all the energy has transformed to elastic potential
energy.
1
Eg = mg!y ; Ee = k(!x)2
2
Solution: The gravitational potential energy as the block is dropped is equal to the elastic
potential energy at full compression of the spring.
Eg = Ee
1
k(!x)2
2
k(!x)2
!y =
2mg
mg!y =
=
(280 N/m)(0.11 m)2
2(0.22 kg)(9.8 m/s 2 )
!y = 0.79 m
This height represents the total distance from where the block was dropped to the lowest
point of the spring, so subtract the maximum compression of the spring, 0.11 m, to get
the height from which the block was dropped.
h = !y " !x
= 0.79 m " 0.11 m
h = 0.68 m
Statement: The block was dropped from a height of 0.68 m.
8. (a) Given: m = 1.0 kg; v = 1.0 m / s; k = 1000.0 N / m
Required: ∆x
Analysis: The kinetic energy of the block transforms fully to elastic potential energy at
the maximum compression of the spring, because the block is at rest at this point.
1
1
Ek = mv 2 ; Ee = k (Δx)2
2
2
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.7-9
Solution: The kinetic energy of the block before it hits the spring is equal to the elastic
potential energy of the spring at its maximum compression.
Ee = Ek
1
1
k(!x)2 = mv 2
2
2
!x =
mv 2
k
(1.0 kg)(1.0 m/s)2
=
1000.0 N/m
!x = 0.032 m
Statement: The maximum compression of the spring is 0.032 m.
(b) The block will travel to the maximum compression of the spring, 0.032 m before
coming to rest.
9. (a) Given: m = 6.0 kg; v = 3.0 m / s; k = 1250 N / m
Required: Δx
Analysis: The kinetic energy of the block transforms fully to elastic potential energy at
the maximum compression of the spring, because the block is at rest at this point.
1
1
Ek = mv 2 ; Ee = k(!x)2
2
2
Solution:
Ee = Ek
1
1
k(!x)2 = mv 2
2
2
!x =
mv 2
k
(6.0 kg)(3.0 m/s)2
1250 N/m
!x = 0.208 m (one extra digit carried)
Statement: The maximum distance the spring is compressed is 0.21 m.
(b) Given: !x = 14 cm = 0.14 m; m = 6.0 kg;vi = 3.0 m / s; k = 1250 N / m
=
Required: vf ; a
Analysis: Some of the kinetic energy of the block transforms to elastic potential energy
as the spring compresses. So the initial kinetic energy transforms to final kinetic energy
and elastic potential energy.
1
1
Ek = mv 2 ; Ee = k(!x)2
2
2
For the acceleration, the only force acting on the block is the spring force, which is equal
to! k∆x away from the spring.
!
F = ma
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.7-10
Solution: The initial kinetic energy is equal to the sum of the final kinetic energy and the
elastic potential energy.
Ek i = Ek f + Ee
1 2 1 2 1
mv = mvf + k(!x)2
2 i
2
2
mvf2 = mvi2 " k(!x)2
vf =
=
mvi2 " k(!x)2
m
(6.0 kg)(3.0 m/s)2 " (1250 N/m)(0.14 m)2
6.0 kg
vf = 2.2 m/s
For the acceleration:
!
!
F = ma
!
k!x [away from the spring] = ma
! k!x [away from the spring]
a=
m
(1250 N/m)(0.14 m) [away from the spring]
=
6.0 kg
!
2
a = 29 m/s [away from the spring]
Statement: When the spring is compressed, the speed of the block is 2.2 m/s, and the
acceleration is 29 m/s2 [away from the spring].
10. Given: k = 440 N / m; !x = 45 cm = 0.45 m; m = 57 g = 0.057 kg; d y = 1.2 m
Required: d x , the horizontal distance
Analysis: Find the speed of the ball as it leaves the machine, and then use projectile
motion equations to determine when it will hit the ground.
Let the y = 0 reference point be the height at which the ball leaves the machine. Thus,
there is no gravitational potential energy at this point. The elastic potential energy of the
spring when it is at its maximum compression transforms to kinetic energy when the ball
leaves the machine.
1
1
Ek = mv 2 ; Ee = k(!x)2
2
2
Once v has been determined, use projectile motion equations to determine the distance,
dx, the ball travels. Since the ball is projected horizontally, the initial launch angle is 0°,
1
and there is no vertical component of velocity: v y = 0 . Use the equation d y = v y t ! gt 2
2
to determine t, and then use the equation d x = vx t to determine d x .
Solution: The elastic potential energy of the spring when it is at its maximum
compression is equal to the kinetic energy when the ball leaves the machine.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.7-11
Ek = Ee
1 2 1
mv = k(!x)2
2
2
k(!x)2
m
v=
(440 N/m)(0.45 cm)2
0.057 g
=
v = 39.5 m/s (one extra digit carried)
Now, use the projectile motion equations.
1
d y = v y t ! gt 2
2
1
d y = ! gt 2
2
t=
!2d y
g
!2(!1.2 m)
9.8 m/s 2
t = 0.495 s (one extra digit carried)
=
d x = vx t
= (39.5 m/s)(0.495 s)
d x = 2.0 ! 101 m
Statement: The horizontal distance that the tennis ball can travel before hitting the
ground is 2.0 ! 101 m.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4.7-12
Chapter 4 Review, pages 214–219
Knowledge
1. (b)
2. (c)
3. (b)
4. (c)
5. (a)
6. (d)
7. (b)
8. True
9. True
10. False. The gravitational potential energy of an object 5 m above the ground in Ontario
is greater than that of an identical object 5 m above the ground on the Moon.
11. False. The joule (J) is the SI unit for two quantities: work and energy.
12. True
13. True
14. False. In an oscillating spring, the elastic potential energy when the spring is
completely compressed is equal to the elastic potential energy when the spring is fully
extended.
Understanding
15. (a) No work is being performed by the car, because it is not moving.
(b) No work is being performed by the snow bank because it is not moving.
16. Answers may vary. Sample answer: Two examples in which a non-zero force acts on
an object, but the total work done by that force is zero, are a person who is pushing an
appliance but is unable to move it; a mass is hanging from a string on a frictionless pulley
attached to a block on a table, but the block is not moving.
17. No, it is not possible to do work on an object if the object does not move. An object
has to move a non-zero distance ∆d for work to be done.
!
18. (a) Given: Fg = 9.31! 103 N [straight downward] ; ! = 4°
!
Required: component of gravitational force parallel to the car’s motion, Fg parallel
!
Analysis: Determine Fg parallel using trigonometry.
Solution:
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4-2
!
!
Fg parallel = Fg sin !
= (9.31" 103 N)sin 4°
!
Fg parallel = 649.4 N [parallel to the slope] (one extra digit carried)
Statement: The component of the gravitational force that acts parallel to the car’s motion
is 649 N. !
(b) Given: Fg parallel = 649.4 N [parallel to the slope] ; !d = 30.0 m
Required: W
Analysis: W = F!d
Solution: W = F!d
= (649.4 N)(30.0 m)
W = 1.95 " 104 J
Statement: The work done on the car by gravity as the car slides is 1.95 ! 104 J.
19. (a) Given: m = 65 kg; !y = 100.0 m
Required: W
Analysis: W002 on 001 = ! Eg ; Eg = mg!y
The work done on 002 by 001 is equal in magnitude but opposite in sign to the work done
by gravity, which is equal to the gravitational potential energy.
Solution: W001 on 002 = ! Eg
= !mg"y
= !(65 kg)(9.8 m/s 2 )(100.0 m)
W001 on 002 = !6.4 # 104 J
Statement: The work done by 001 on 002 is −6.4 ! 104 J.
(b) Given: m = 65 kg; ∆y = 100.0 m
Required: W
Analysis: The work done by gravity is equal to the gravitational potential energy.
W = Eg ; Eg = mg!y
Solution: W = Eg
= mg!y
= (65 kg)(9.8 m/s 2 )(100.0 m)
W = 6.4 " 104 J
Statement: The work done by gravity on 002 is 6.4 ! 104 J.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4-3
(c) Given: m = 65 kg; v = 2.5 m / s
Required: Ek
1
2
1
Solution: Ek = mv 2
2
1
= (65 kg)(2.5 m/s)2
2
Ek = 2.0 ! 102 J
Analysis: Ek = mv 2
Statement: Spy 002’s kinetic energy is 2.0 ! 102 J.
(d) Spy 002’s gravitational potential energy when she leaves the top of the building is
6.4 !104 J , and when she reaches the ground, it is 0. Thus, the change in gravitational
potential energy is –6.4 ! 104 J.
20. Work is the energy required when a force moves an object a certain distance in the
same direction as the force.
21. Given: m = 68 kg; v1 = 5.8 m / s; v2 = 6.9 m / s
Required: W
1
Analysis: Ek = mv 2
2
The work done is equal to the difference in kinetic energy.
Solution: W = Ek 2 ! Ek 1
1 2 1 2
mv ! mv
2 2 2 1
1
= m(v22 ! v12 )
2
1
= (68 kg)[(6.9 m/s)2 ! (5.8 m/s)2 ]
2
W = 470 J
Statement: The work that the sprinter does is 4.7 ! 102 J.
22. Given: m = 20.0 kg; vi = 2.0 m / s; vf = 0 m / s
Required: Wf
Analysis: The work done by friction is equal to the opposite of the change in kinetic
energy, since friction works against the curling stone.
1
W = !"Ek ; Ek = mv 2
2
=
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4-4
Solution: W = !"Ek
= !(Ek 2 ! Ek 1 )
#1
&
1
= ! % mv22 ! mv12 (
2
$2
'
1
= ! mv22
2
1
= ! (20.0 kg)(2.0 m/s)2
2
W = !4.0 ) 101 J
Statement: The work done on the stone by friction is −4.0 ! 101 J.
23. Given: m = 60.0 kg; !y = 10.0 m
Required: ΔEg
Analysis: Let the y = 0 reference point be the surface of the water. The change in
gravitational potential energy is the negative of the gravitational potential energy as the
diver leaves the diving board.
Eg = mg Δy
Solution: !Eg = " Eg
= "mg!y
= "(60.0 kg)(9.8 m/s 2 )(10.0 m)
!Eg = "5.9 # 103 J
Statement: The change in gravitational potential energy is −5.9 × 103 J .
24. Given: !y = 1.2 m; !Eg = 5.8 J
Required: m
Analysis: Let the y = 0 reference point be the floor. The change in gravitational potential
energy is equal to the gravitational potential energy at the level of the shelf.
!Eg = Eg = mg!y
Solution: !Eg = mg!y
m=
=
!Eg
g!y
5.8 J
(9.8 m/s 2 )(1.2 m)
m = 0.49 kg
Statement: The mass of the case of cereal is 0.49 kg.
25. Both mass and height are factors in gravitational energy, and mass is proportional to
volume. So volume and height of the waterfall are important for hydroelectric power
generation.
26. The energy increases, because height is a factor in gravitational potential energy. So
when the height increases, the gravitational potential energy increases.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4-5
27. Answers may vary. Sample answer:
Work
Description the amount of energy required to move an
object a certain distance in the direction of an
applied force
Units
joules (J)
Example
A worker pushes a crate up a ramp.
28. Given: !y = 1.2 m; m = 55 g = 0.055 kg
Required: Eg
Analysis: Let the y = 0 reference point be the ground.
Eg = mg Δy
Power
the amount of work done
over a certain amount of
time
watts (W)
A sled dog pulls a sled to
win a 100-m race in 10 s.
Solution: Eg = mg!y
= (0.055 kg)(9.8 m/s 2 )(1.2 m)
Eg = 0.65 J
Statement: The gravitational potential energy of the egg before it falls is 0.65 J.
29. Given: t = 1 s; m = 5.7 ! 105 kg; "y = 21 m
Required: P
W
Analysis: P = ; W = mg!y
t
W
Solution: P =
t
mg!y
=
t
(5.7 " 105 kg)(9.8 m/s 2 )(21 m)
=
1s
8
P = 1.2 " 10 W
Statement: The power generated is 1.2 ! 108 W.
30. Answers may vary. Sample answer: A skier going down a hill converts gravitational
potential energy to kinetic energy and thermal energy (due to friction with the hill).
31. Answers may vary. Sample answer: A short diving board would be stiffer than a
longer diving board. When a person is standing on a short board, the vertical
displacement of the board from its equilibrium position will be less than the vertical
displacement of a longer board with the same person standing on it. If the force exerted
on both boards is the same, but the vertical displacement differs, then the smaller the
vertical displacement, the greater the spring constant. The spring constant is a constant of
proportionality. Material that is stiff will need a greater force to extend or compress it.
Therefore, the smaller diving board will have a larger spring constant.
32. Answers may vary. Sample answer: The elastic potential energy of a spring is
proportional to the square of the compression or extension distance of the spring. The
proportionality constant is k. If k is large, the spring is stiff. If k is small, the spring is
loose.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4-6
33. Answers may vary. Sample answer: The mass on the spring with the smaller spring
constant will oscillate more quickly than the mass on the spring with the larger spring
constant.
34. Answers may vary. Sample answer: The oscillation of the larger mass will have a
larger amplitude than that of the smaller mass.
35. Answers may vary. Sample answer: The elastic potential energy is proportional to the
distance the spring (in this case the bowstring) is stretched. The kinetic energy when the
arrow is released is equal to the elastic potential energy. The kinetic energy is
proportional to the square of the speed. Thus, the more the bowstring is stretched, the
greater the elastic potential energy, the greater the kinetic energy, and the greater the
speed.
36. Answers may vary. Sample answer: In the kinetic energy equation, speed is inversely
proportional to the square root of mass. Thus, the lighter the mass, the greater the speed.
37. Given: !x; k; m
Required: v
Analysis: The kinetic energy of the ball as it leaves the spring is equal to the elastic
potential energy when the spring is compressed:
1
1
Ek = Ee ; Ek = mv 2 ; Ee = k(!x)2
2
2
Solution:
Ek = Ee
1 2 1
mv = k(!x)2
2
2
v2 =
k(!x)2
m
k(!x)2
m
" k%
v=$
' !x
# m&
v=
! k$
Statement: The ball’s initial speed is v = #
& 'x .
" m%
38. There is no work done on the object by the force. Work requires the object to move in
the direction of the applied force.
Analysis and Application
39. Given: ! = 18.5°; W = 214 J
Required: !d
Analysis: Draw an FBD of the toboggan. Use trigonometry to determine the applied
force in the direction of the movement of the sled.
W = F!d
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4-7
Solution:
Fx = F cos θ
W = Fx !d
W = F cos" (!d)
W
F cos"
214 J
=
(11.8 N)cos18.5°
!d =
!d = 19.1 m
Statement: The sled moves 19.1 m.
40. No. You go flying off the merry-go-round in a direction perpendicular to centripetal
force. Since the angle is 90°, and cos 90° = 0 , the work done is 0.
41. (a) In the diagram, the lettered arrows represent the direction of motion of the comet,
and the dashed arrows represent the gravitational force of the Sun. At points A and D, the
motion of the comet is perpendicular to the gravitational force of the Sun, so the Sun does
no work on the comet. At points B and C, there is a component of the gravitational force
of the Sun that is parallel and opposite to the motion of the comet, so the Sun does
negative work on the comet. At points E and F, there is a component of the gravitational
force of the Sun that is parallel and in the same direction as the motion of the comet, so
the Sun does positive work on the comet.
(b) As the comet approaches the Sun, the work done by gravity decreases to zero, in a
direction toward the Sun, in the same direction as the comet.
(c) As the comet moves away from the Sun, the work done by gravity increases from
zero, but in a direction toward the Sun, opposite the motion of the comet.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4-8
42. The units of force are newtons (N), or kilograms times metres per second squared
" kg ! m %
(kg·m/s2). The units of distance are metres (m). So the units of work are $ 2 ' ![m] or
# s &
" kg ! m 2 %
$ 2 ' . The units of mass are kilograms (kg), and the units of speed are metres per
# s
&
2
" kg ! m 2 %
"m %
second (m/s). So the units of kinetic energy are [kg]! $ ' or $ 2 ' . The units of
#s&
# s
&
work done by a constant force and kinetic energy are the same.
43. We are not given the mass of the motorcycle. A reasonable value for the horizontal
applied force needed to move the motorcycle is 100 N. Draw an FBD of the motorcycle.
Using trigonometry,
Fx
= cos!
Fa
Fx
cos!
100 N
Fa =
cos!
Use graphing software to graph this equation.
Fa =
As the angle increases, the applied force required to move the motorcycle increases.
44. Given: m1 = 1000 kg; m2 = 2000 kg; m3 = 3000 kg; m4 = 4000 kg; v = 10 m / s;
!d = 1 km = 1000 m
Required: F
Analysis: The work done by the spacecraft on each load is equal to the kinetic energy of
the load.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4-9
1 2
mv
2
Solution: Solve the equation W = Ek for F, and complete a table of values for the four
values of m. Then, draw a scatter plot graph of the data.
W = Ek
W = F!d ; Ek =
1 2
mv
2
mv 2
F=
2!d
m(10 m/s)2
=
2(1000 m)
F!d =
F=
Mass, m
1000 kg
2000 kg
3000 kg
4000 kg
m
m/s 2
20
Force,
F=
m
m / s2
20
50 N
100 N
150 N
200 N
Statement: The force, in newtons, is one twentieth the mass, in kilograms. The force and
the mass are linearly related.
45. Given: m = 2000 kg;v1 = 10 m / s;v2 = 15 m / s;v3 = 20 m / s;v4 = 25 m / s;
!d = 1 km = 1000 m
Required: F
Analysis: The work done by the spacecraft on each load is equal to the kinetic energy of
the load.
1
W = F!d ; Ek = mv 2
2
Solution: Solve the equation W = Ek for F, and complete a table of values for the four
values of v. Then, draw a scatter plot graph of the data.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4-10
W = Ek
1 2
mv
2
mv 2
F=
2!d
(2000 kg)v 2
=
2(1000 m)
F!d =
F = v 2 kg/m
Speed, v Force, F = v 2 kg / m
10 m/s
100 N
15 m/s
225 N
20 m/s
400 N
25 m/s
625 N
Statement: The force, in newtons, is the square of the speed, in metres per second. The
force and the speed are quadratically related.
46. (a) The student’s velocity is changing, because the direction is changing.
(b) The student’s speed is constant. There is no friction, so the initial speed does not
change.
(c) The student’s kinetic energy is constant, since the student’s speed is constant, and
kinetic energy is proportional to the square of speed.
47. The value of g decreases as we move farther from Earth’s surface.
48. (a) Use the value of g for the Moon instead of that for Earth.
(b) Answers may vary. Sample answer:
You could use a motion sensor to measure the speed an object attains when dropped from
a certain height. If the mass is m, the speed is v, and the change in height is Δy , then
Eg = Ek
1
mv 2
2
v2
g=
2!y
mg!y =
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4-11
Use the height and the value of v from the motion detector to determine g on the Moon.
Do this experiment several times with different height and speed values, and determine
the average value of g.
49. (a) Since gravitational potential energy is proportional to height, if the height doubles,
the energy doubles. So plant X generates twice as much power as plant Y.
(b) Gravitational potential energy is also proportional to mass, which is proportional to
volume, so if the volume doubles, the energy doubles. So plants X and Y now generate
the same amount of power.
50. Given: m = 1.3 kg; vi = 1.8 m / s; vf = 0.9 m / s; !y = 4.0 m
Required: energy lost through air resistance and friction
Analysis: The energy at the top of the chute is a combination of kinetic energy and
gravitational potential energy. The energy when the package reaches the floor is all
kinetic energy. The difference in energy between that at the top of the chute and that at
the floor is the energy lost through air resistance and friction.
1
Eg = mg!y ; Ek = mv 2
2
Solution: Let Elost represent the energy lost.
Elost = Eg top + Ek top ! Ek floor
1
1
= mg"y + mvi2 ! mvf2
2
2
#
1
1 &
= m % g"y + vi2 ! vf2 (
2
2 '
$
)
,
1
1
= (1.3 kg) +(9.8 m/s 2 )(4.0 m) + (1.8 m/s)2 ! (0.9 m/s)2 .
2
2
*
Elost = 53 J
Statement: The energy lost through air resistance and friction is 53 J.
1
51. Given: m1 = m2 ; !y2 = !y1
4
Required: comparison of v1 to v2
Analysis: The gravitational potential energy when the balloons are dropped is equal to
the kinetic energy when they hit the ground.
1
Eg = mg!y ; Ek = mv 2
2
Solution:
Eg 1 = Ek 1
1
m1 v12
2
v1 = 2 g Δy1
m1 g Δy1 =
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4-12
Eg 2 = Ek 2
m2 g!y2 =
1
m2 v22
2
v2 = 2g!y2
"1
%
= 2g $ !y1 '
#4
&
=
v2 =
1
(2g!y1 )
4
1
2g!y1
2
Statement: The speed of the blue balloon is half the speed of the red balloon.
52. The pole vaulter’s kinetic energy, which is proportional to the square of his speed, as
he approaches the jump transforms into the gravitational potential energy that will help
him to clear the bar. Thus, approach speed is important to the pole vaulter.
53. Given: m = 57 kg; !y1 = 45 m; !y2 = 25 m
Required: v
Analysis: The skier’s total energy is the same at the top of the first peak as at the top of
the second peak. At the top of the first peak, her energy is all gravitational potential
energy, because she starts from rest. At the top of the second peak, her energy is a
combination of gravitational potential energy and kinetic energy.
1
Eg = mg!y ; Ek = mv 2
2
Solution: Solve for v in the equation Eg 1 = Eg 1 + Ek .
Eg 1 = Eg 2 + Ek
mg!y1 = mg!y2 +
1
mv 2
2
1 2
v = g(!y1 " !y2 )
2
v = 2g(!y1 " !y2 )
= 2(9.8 m/s 2 )(45 m " 25 m)
v = 2.0 # 101 m/s
Statement: The skier’s speed at the second peak is 2.0 ! 101 m/s.
54. Given: m = 450 kg; v = 3.5 m / s; !x = 2.0 m
Required: k
Analysis: The kinetic energy at the moment the car hits the spring is transformed
completely to elastic potential energy when the spring is fully compressed.
1
1
Ek = mv 2 ; Ee = k(!x)2
2
2
Solution: The kinetic energy at the moment the car hits the spring is equal to the elastic
potential energy when the spring is fully compressed.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4-13
Ee = Ek
1
1
k(!x)2 = mv 2
2
2
mv 2
k=
(!x)2
=
(450 kg)(3.5 m/s)2
(2.0 m)2
k = 1400 N/m
Statement: The spring constant is 1.4 ! 103 N/m.
55. (a) Given: vi = 27 m / s; ! = 20°;m = 0.43 kg; g = 9.8 m / s 2
Required: Δy , the maximum height of the ball
Analysis: Determine the y-component of the initial velocity using trigonometry.
At the maximum height of the ball, the y-component of the velocity is zero, so the kinetic
energy is zero. The kinetic energy when the ball is kicked is equal to the potential energy
at the maximum height.
1
2
Eg = mg Δy ; Ek = mv2
Solution: The y-component of the initial velocity is visinθ .
Eg = Ek
1
mv 2
2
v2
!y =
2g
mg!y =
=
[(27 m/s)sin 20°]2
2(9.8 m/s 2 )
!y = 4.4 m
Statement: The maximum height of the ball is 4.4 m.
(b) Since there is no air resistance, the speed of the ball when it lands is equal to the
speed of the ball when it was kicked. Thus, the speed of the ball when it lands is 27 m/s.
56. (a) Given: m = 55 kg; Δy = 1.3 m
Required: Wg
Analysis: The work done by gravity is the negative of the gravitational potential energy.
Wg = − Eg ; Eg = mg Δy
Solution: Wg = ! Eg
= !mg"y
= !(55 kg)(9.8 m/s 2 )(1.3 m)
Wg = !701 J (one extra digit carried)
Statement: The work done by gravity is −7.0 ! 102 J.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4-14
(b) Given: m = 55 kg; vi = 5.4 m / s; !y = 1.3 m
Required: vf
Analysis: The student’s total energy when she leaves the trampoline’s surface is all
kinetic energy. As she rises in the air, the kinetic energy transforms into gravitational
potential energy. If she has not reached her maximum height, she will still have some
kinetic energy.
1
Ek = mv 2 ; Eg = mg!y
2
Solution:
Ek i = Eg + Ek f
1
1
mvi2 = mg!y + mvf2
2
2
1 2 1 2
v = v " g!y
2 f 2 i
vf2 = vi2 " 2g!y
vf = vi2 " 2g!y
= (5.4 m/s)2 " 2(9.8 m/s 2 )(1.3 m)
vf = 1.9 m/s
Statement: The student’s speed at 1.3 m above the trampoline is 1.9 m/s.
(c) No, she has not reached her maximum height because she still has kinetic energy. At
the maximum height, there is no kinetic energy.
57. (a) Given: k = 4.5 N / m; !x1 = 0.75 m; !x2 = 0.5 m
Required: Elost
Analysis: The amount of energy lost is equal to the change in elastic potential energy.
Ee =
1
k (!x) 2
2
Solution: Elost = Ee 1 ! Ee 2
1
1
k("x1 )2 ! k("x2 )2
2
2
1
= k[("x1 )2 ! ("x2 )2 ]
2
1
= (4.5 N/m)[(0.75 m)2 ! (0.50 m)2 ]
2
= 0.703 J (two extra digits carried)
=
Elost = 0.7 J
Statement: The system has lost 0.7 J of energy.
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Chapter 4: Work and Energy
4-15
(b) Given: Elost = 0.703 J; t = 15 min = 900 s
Required: P
Analysis: The work done by the damping is equal to the energy lost.
W
P=
t
W
Solution: P =
t
E
= lost
t
0.703 J
=
900 s
P = 7.8 ! 10"4 W
Statement: The power loss of the system is 7.8 ! 10−4 W.
58. Given: k = 6.0 N / m; !x = 0.40 m
Required: m
Analysis: Let the y = 0 reference point be the point where the ball begins its bounce
back up. Therefore, Δy = Δx . The gravitational potential energy when the ball is released
transforms to elastic potential energy at the point where the ball bounces.
1
2
Eg = mg Δy; Ee = k (Δx)2
Solution:
Eg = Ee
1
k(!x)2
2
k(!x)2
m=
2g!y
mg!y =
=
=
(6.0 N/m)(0.40 m)2
2(9.8 m/s 2 )(0.40 m)
(6.0 kg/ s 2 )(0.40 m ) 2
2(9.8 m/s 2 )(0.40 m )
m = 0.12 kg
Statement: The mass of the ball is 0.12 kg.
59. (a) Given: m = 0.50 kg; !x1 = 0.25 m; v = 1.5 m/s
Required: k
Analysis: The elastic potential energy when the ball is released transforms to kinetic
energy at the maximum speed of the ball.
1
1
Ee = k(!x)2 ; Ek = mv 2
2
2
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Chapter 4: Work and Energy
4-16
Solution:
Ee = Ek
1
1
k(!x1 )2 = mv 2
2
2
k=
=
mv 2
(!x1 )2
(0.50 kg)(1.5 m/s)2
(0.25 m)2
k = 18 N/m
Statement: The spring constant is 18 N/m.
(b) Given: !x1 = 0.25 m; !x2 = 0.125 m; k = 18 N / m; m = 0.50 kg
Required: v
Analysis: When the ball is halfway to its equilibrium point, its total energy consists of
kinetic energy and elastic potential energy. At its equilibrium point, the total energy
consists only of elastic potential energy.
1
1
Ee = k(!x)2 ; Ek = mv 2
2
2
Solution:
Ee 1 = Ee 2 + Ek
1
1
1
k(!x1 )2 = k(!x2 )2 + mv 2
2
2
2
mv 2 = k(!x1 )2 " k(!x2 )2
v=
=
k(!x1 )2 " k(!x2 )2
m
(18 N/m)(0.25 m)2 " (18 N/m)(0.125 m)2
0.50 kg
v = 1.3 m/s
Statement: The speed of the ball halfway to its equilibrium point is 1.3 m/s.
(c) Given: v = 1.3 m / s; m = 0.50 kg; k = 18 N / m; !x = 0.25 m
Required: fraction of energy converted from elastic potential energy to kinetic energy
Analysis: Determine the fraction of Ek over Ee 1 from (b).
1
1
Ek = mv 2 ; Ee = k (Δx)2
2
2
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Chapter 4: Work and Energy
4-17
1 2
mv
Ek
= 2
Solution:
Ee 1
k(!x1 )2
2
mv 2
=
k(!x1 )2
=
(0.50 kg)(1.3 m/s)2
(18 N/m)(0.25 m)2
Ek
= 0.75
Ee
Statement: The fraction of energy converted from elastic potential energy to kinetic
energy is
3
.
4
60. The gravitational potential energy of the heavier object at the equilibrium point is
greater than that of the lighter object, so the elastic potential energy at the maximum
stretch of the heavier object will be greater, since the gravitational potential energy equals
the elastic potential energy. Since elastic potential energy is proportional to the square of
the maximum stretch amount, if the elastic potential energy is greater, the maximum
stretch amount will be greater.
61. At the beginning of the jump, the jumper’s gravitational potential energy is at a
maximum. As the jumper falls, the gravitational potential energy decreases and the
kinetic energy increases. At the full length of the bungee cord before it starts stretching,
the kinetic energy is at a maximum and the gravitational potential energy continues to
increase. As the bungee cord stretches, the elastic potential energy increases from zero up
to its maximum when the cord is fully stretched. When the cord is fully stretched, the
elastic potential energy is at its maximum, and gravitational potential energy and kinetic
energy are zero. As the jumper bounces back up, the elastic potential energy decreases
and the kinetic and gravitational potential energy increase.
62. Answers may vary. Sample answer:
thermal energy: When a skater stops on the ice, kinetic energy transforms to thermal
energy through friction between the skate blades and the ice.
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Chapter 4: Work and Energy
4-18
63. (a)
(b) The kinetic energy of the car is equal to the work done by the road on the car to stop
the car. Therefore,
Ek = W
1 2
mv = F!d
2
mv 2
!d =
2F
Since the force, F, and the mass, m, are constant, the stopping distance, ∆d, is
proportional to the square of the speed, v. The graph is quadratic.
(c) Answers may vary. Sample answer: The data shows that the stopping distance is
proportional to vehicle speed, so the faster you are driving, the greater the distance
required to stop the vehicle. Therefore, it is important to make sure that there is enough
distance between your car and the car in front of you so that you can stop safely. This
data also proves that tailgating is dangerous and should be avoided.
Evaluation
64. Answers may vary. Sample answer: The gravitational force is the force exerted by
gravity on the crate due to its mass, and it is directed downwards. The normal force is the
force between two objects that are in contact, and its direction is perpendicular to the
surface. The applied force is the force the worker applies to push the crate, and its
direction is up the ramp. Friction opposes the applied force, so its direction is down the
ramp. The gravitational force, the applied force, and friction do work. The normal force
does zero work. This is because the normal force is perpendicular to the direction of the
displacement. The applied force does work because it is in the direction of the
displacement, and friction does negative work because it opposes the direction of the
displacement. The gravitational force does work on the crate as well because when it is
separated into component vectors, one of the components does negative work.
65. Answers may vary. Sample answers:
(a) If Earth were less massive, the force of gravity would be less, so the Moon would not
be attracted as strongly. Tides would be different. Birds would fly more easily. It would
not take as much fuel to fly an airplane.
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4-19
(b) If Earth were more massive, the force of gravity would be greater, so the Moon would
be attracted more strongly. This would affect the same things as in (a), but in the opposite
way.
66. Answers may vary. Sample answer: The gravitational potential energy of the spring
toy at the top of the stairs is transformed to kinetic energy when the toy is started down
the stairs. The energy is passed from coil to coil in the spring as it moves down the stairs.
67. Answers may vary. Sample answer: Roller coasters take advantage of gravitational
potential energy to start moving down hills. The higher the hill, the faster the roller
coaster car will move at the bottom of the hill, and the more easily it will make it up the
next hill, and so on.
68. Answers may vary. Sample answer: As the frame of the car crumples, it absorbs the
energy from the collision, so less of the impact is felt by the people in the car.
Reflect on Your Learning
69. Answers may vary. Sample answer: I found the concept of work difficult to
understand since work is done only if a displacement occurs. If I push on a stationary
object like a wall, I exert a force or energy pushing against the wall, but no work is done
because the wall does not move. To understand this concept, I would have to search for
many examples of work being done versus no work being done to keep this concept clear.
Another concept that I found difficult to understand is that the direction of the spring
force is opposite to the displacement of the spring from its equilibrium position. If you
stretch as a spring upwards so that the displacement from the equilibrium position is
upwards, the spring force is actually downwards. To understand this concept fully, it
would help if I could compress and stretch some springs to see how the forces work in
real life.
70. After having read the chapter, I have a better understanding of the different sources of
commercial energy, especially hydroelectric power. While hydroelectric power is cleaner
than burning fossil fuels for electricity generation, it still has a huge environmental
impact. Water resources are diminished, land use is diverted for power dam construction,
wildlife habitats are destroyed, and the natural path of water sources are artificially
diverted, changing ecosystems. While I still prefer this over fossil fuel–burning plants,
alternative sources for electricity generation, such as solar and wind power, should be
investigated further.
71. Answers may vary. Sample answer: One topic that I am still unsure of is the use of
components in determining work. What confuses me is which angle and trigonometric
ratio I should use to calculate the component vectors and when I should be separating the
components. One way I can improve my understanding is to do more questions involving
these components such as questions involving ramps. Another way for me to improve my
understanding is to draw vector diagrams, including right angle triangles to help me
visualize these problems.
72. Answers may vary. Sample answer: One instance where I can apply what I learned in
this chapter is when I am shovelling snow. Since I know that work is done only when
there is a displacement, I should decrease the angle at which I apply the force on my
shovel so that the work is done more efficiently. Another instance is when I am at the
swimming pool. To make little to no splash when I am diving, I need to pick a diving
board that is long and loose so that its spring constant is small. This means that I will to
Copyright © 2012 Nelson Education Ltd.
Chapter 4: Work and Energy
4-20
use less force to go up a greater distance. This will give me more time in the air to
execute my dive.
Research
73. Answers may vary. Sample answer: One famous scientist who used pendulums to
study gravity and the orbital motions of the plant was Galileo Galilei. He formulated his
Law of Falling Bodies by studying the motions of pendulums as well as objects rolling
and sliding down inclines. He theorized that all bodies would fall with the same
acceleration due to gravity, which he calculated to be 32 ft/s or 9.75 m/s. This is very
close to the value 9.81 m/s, the accepted value of g in today’s society.
Leon Foucault proved that Earth rotates by using a pendulum tied to a building. A
pendulum works because it is affected by gravity and inertia. As a pendulum is pulled
back and then released, it swings down because of gravity and swings up because of
inertia. As the pendulum swings, it seems to be rotating. However, since the way it is
suspended prevents it from rotating, it must be the building that is rotating. Since the
building is attached to Earth, it must be Earth that is rotating.
A simple experiment that can be done to determine the value of g using a
pendulum.
1. Set up a pendulum so that it swings from a fixed end. For example, a ring stand and a
pendulum clamp can be used.
2. Pull back the pendulum and release it making sure that there is very little sideways
motion involved.
3. Determine the period of the pendulum, T, by recording the time it takes for the
pendulum to swing a set number of times and dividing the time by that number.
4. Repeat this several times with pendulums of different lengths, L.
5. Graph the T2 v. L using the data collected.
6. From the rearrangement of the equation below, g can be determined.
T = 2!
L
g
" L%
T 2 = 4! 2 $ '
# g&
T 2 4! 2
=
L
g
4! 2
slope =
g
g=
4! 2
slope
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Chapter 4: Work and Energy
4-21
74. Answers may vary. Sample answer:
Hydroelectric Plant
Advantages
Disadvantages
Churchill,
• natural basin so no need to
• long distance transmissions
New Brunswick
construct dams
for electricity to be used
• brings economic
• remote location and harsh
opportunities to New
climate
Brunswick and Labrador
• alters natural beauty of
Churchill Falls and
• provides enough electricity
to power three cities the
surrounding area
size of Montreal
Niagara Falls,
• supplies one quarter of
• not all of the power goes
Ontario
power used in Ontario and
directly to Ontario;
New York State
Canadians are able to draw
56 500 cubic feet of water
• hydroelectric power may be
per second and Americans
sold between generating
are allowed 32 500 cubic
stations in Ontario and
feet of water per second
generating stations in New
through international
York state when needed
agreement
• constraints made to limit
•
issues with the
the amount of water used
environment such as ice
for industry because of
flows and flooding
tourism
• regulating flow of water
slows down erosion so
prolongs life of Niagara
Falls
75. Answers may vary. Sample answer: The physics of carousels is based on centripetal
force. Centripetal force is the force that keeps the horses and people moving in a circular
motion. In carousels, this centripetal force is supplied by the platform that supports the
carousel. The carousel has to be at a low speed so that centrifugal force does not get
strong enough to take over the centripetal force that is acting on the people riding the
carousel. The centrifugal force is not an actual force. It is when centripetal force stops
working and a body’s inertia takes over. Since inertia is the resistance of a body to its
change in motion, the rider would move in a straight line that is tangent to the centripetal
force.
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Chapter 4: Work and Energy
4-22
Chapter 4 Self-Quiz, page 213
1. (b)
2. (b)
3. (b)
4. (d)
5. (c)
6. (d)
7. (c)
8. False. Work is a scalar quantity.
9. False. Work can be a negative quantity.
10. True
11. True
12. False. The production of hydroelectricity provides fairly clean energy, but there are
still environmental concerns.
13. False. There are no examples of isolated systems in the real world.
14. True
15. True
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Chapter 4: Work and Energy
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