REVIEW AND SYNTHESIS: CHAPTERS 6–8
Review Exercises
1. (a) Strategy Multiply the extension per mass by the mass to find the maximum extension required.
Solution
⎛ 1.0 mm ⎞
⎛ 1000 g ⎞ ⎛ 1 m ⎞
⎜
⎟ (5.0 kg) ⎜
⎟⎜
⎟ = 0.20 m
⎝ 25 g ⎠
⎝ 1 kg ⎠ ⎝ 1000 mm ⎠
(b) Strategy Set the weight of the mass equal to the magnitude of the force due to the spring scale. Use
Hooke’s law.
Solution
Weight = mg = kx, so k =
mg (5.0 kg)(9.80 N kg)
=
= 250 N m .
x
0.20 m
2. Strategy Plot force on the y-axis and the spring length on the x-axis. Use the graph to answer the questions.
Solution Graph the data.
Force, F (N)
1.500
1.000
0.500
0.000
10.0
15.0
20.0
25.0
Spring length, x (cm)
(a) Determine the slope of the line to find k, since F = kx.
1.20 N − 0 N
1.20 N
=
= 0.15 N cm
k=
20.0 cm − 12.0 cm 8.0 cm
(b) The force on the spring is zero when the spring is relaxed, so from the figure, x0 = 12 cm .
3. Strategy Use conservation of energy and Newton’s second law.
Solution Relate the speed to the length of the cord.
1
1
L
∆K = mv 2 − 0 = mv 2 = −∆U = −mg ∆y = mg , so v 2 = gL.
2
2
2
Use Newton’s second law and solve for the tension.
v2
gL
∑ Fy = T − mg = mar = m
=m
= mg , so T = 2mg .
r
L
435
Review and Synthesis: Chapters 6–8
Physics
4. Strategy The work done by the muscles is 22% of the energy expended. The gravitational potential energy
gained by the person is equal to the work done by the muscles.
Solution 0.22 E = W = ∆U = mgh, so
mgh (80.0 kg)(9.80 N kg)(15 m)
E=
=
= 53 kJ .
0.22
0.22
5. (a) Strategy Use the conservation of energy.
Solution Find the work done by friction.
N
1
Wtotal = Wfriction + Wgrav = Wfriction + mgd sin θ = ∆K = 0 − mvi 2 , so
2
1
1
⎛
⎞
Wfriction = − mvi 2 − mgd sin θ = − m ⎜ vi 2 + gd sin θ ⎟
2
⎝2
⎠
⎡1
⎤
= −(100 kg) ⎢ (2.00 m s)2 + (9.80 m s 2 )(1.50 m) sin 30.0° ⎥
⎣2
⎦
= −940 J.
fk
d
mg
y
θ
mg cos θ
mg sin θ
x
θ
Thus, the energy dissipated by friction was 940 J .
(b) Strategy Use Newton’s second law.
Solution Find the normal force on the crate.
ΣFy = N − mg cos θ = 0, so N = mg cos θ .
Since vfx 2 − vix 2 = 0 − vi 2 = 2a x ∆x = 2a x d , the acceleration of the crate is − vi 2 (2d ).
Find the force of sliding friction.
v2
ΣFx = − f k + mg sin θ = − µk mg cos θ + mg sin θ = ma x = − m i , so
2d
vi 2
(2.00 m s)2
µk = tan θ +
= tan 30.0° +
= 0.734 .
2dg cos θ
2(1.50 m)(9.80 m s 2 ) cos 30.0°
6. Strategy Use conservation of energy.
Solution Find the speed of the packing carton at the bottom of the inclined
plane.
1
1
∆K = mvf 2 − mvi 2 = −∆U = −mg ∆y, so
2
2
2.0
m
30.0°
vf = vi 2 − 2 g ∆y = (4.0 m s)2 − 2(9.80 m s 2 )[0 − (2.0 m) sin 30.0°]
= 6.0 m s .
7. Strategy Use conservation of energy.
Solution Find the maximum height of the swing.
1
v2
(6.0 m s) 2
∆K = 0 − mv 2 = −∆U = mghi − mghmax , so hmax =
+ hi =
+ 0.50 m = 2.3 m .
2
2g
2(9.80 m s 2 )
436
(2.0 m) sin θ
Physics
Review and Synthesis: Chapters 6–8
8. Strategy Use conservation of energy and Newton’s second law.
Solution Find the normal force on the crate.
ΣFy = N − mg cos θ = 0, so N = mg cos θ .
fk
N
θ
Find the force of sliding friction.
ΣFx = − f k + mg sin θ = − µk mg cos θ + mg sin θ = ma x , so
y
a x = − µk g cos θ + g sin θ = −0.70(9.80 m s 2 ) cos 53° + (9.80 m s2 ) sin 53° = 3.7 m s2 .
x
Therefore, the acceleration of the block is 3.7 m s 2 down the ramp .
mg cos θ
mg sin θ
mg
θ
9. Strategy The collision is inelastic. Use conservation of momentum and energy.
Solution Write equations using conservation of momentum and energy.
momentum: mvi = (m + M )vf
1
energy: (m + M )vf2 = (m + M ) g ∆y
2
Find the initial speed of the putty.
2
1
⎛ mvi ⎞
(m + M ) ⎜
⎟ = ( m + M ) g ∆y
2
⎝m+M ⎠
2
⎛ m ⎞ 2
⎜
⎟ vi = 2 g ∆y
⎝m+M ⎠
2
2
⎛ 0.50 kg + 2.30 kg ⎞
⎛m+M ⎞
2
vi = 2 g ∆y ⎜
⎟ = 30 m s
⎟ = 2(9.8 m s )(1.50 m) ⎜
0.50 kg
m
⎝
⎠
⎝
⎠
10. Strategy Use conservation of energy. The rotational inertia of a hollow cylinder is I = mr 2 .
Solution Find d, the distance the cylinder travels up the incline.
2
1
1
1
1
⎛v ⎞
0 = ∆K + ∆U = − mvi 2 − I ω 2 + mgd sin θ = − mvi 2 − mr 2 ⎜ i ⎟ + mgd sin θ ,
2
2
2
2
⎝r⎠
2
2
v
(3.00 m s)
=
= 1.53 m .
so d = i
g sin θ (9.80 m s 2 ) sin 37.0°
vi
d
r
y
θ
x
11. Strategy The rotational inertia of a wheel about its central axis is I = 12 MR 2. Use the rotational form of
Newton’s second law.
Solution
(a) I =
1
1
MR 2 = (20.0 kg)(0.224 m)2 = 0.502 kg ⋅ m 2
2
2
(b) The torque required to overcome the friction must be added to that necessary to accelerate the wheel to 1200
rpm in 4.00 s in the absence of friction to get the net torque necessary to accelerate the wheel to 1200 rpm in
4.00 s. Find the torque.
⎡ ∆ω ⎛ ∆ω ⎞ ⎤
1
Στ = I α + I α f = I (α + α f ) = MR 2 ⎢
+⎜
⎟ ⎥
2
⎣ ∆t ⎝ ∆t ⎠f ⎦
=
1
⎛ 1200 rpm 1200 rpm ⎞ ⎛ 2π rad ⎞⎛ 1 min ⎞
+
(20.0 kg)(0.224 m)2 ⎜
⎟ = 17 N ⋅ m
2
60.0 s ⎟⎠ ⎜⎝ rev ⎟⎜
⎝ 4.00 s
⎠⎝ 60 s ⎠
437
Review and Synthesis: Chapters 6–8
Physics
12. Strategy Use the work-kinetic energy theorem. The rotational inertia of a thin hoop is I = mr 2 . The distance d
the bike travels while slowing is equal to the distance the friction force is applied to each wheel.
Solution The force of friction on one of the wheels due to one brake pad is f k = µk N = 0.90 N . Assuming
constant acceleration, the distance the bike travels in the time ∆t = 4.5 s is ∆x = d = (1 2)(vfx + vix )∆t = (1 2)vi ∆t.
Find the normal force on the wheel due to one brake pad.
1
⎛ v ∆t ⎞
⎛1
⎞
Wtotal = Wfriction = −4 f k d = −4(0.90) N ⎜ i ⎟ = −1.8 Nvi ∆t = ∆K = 0 − mb vi 2 − 2 ⎜ I w ω 2 ⎟
2
⎝2
⎠
⎝ 2 ⎠
2
1
1
⎛v ⎞
= − mb vi 2 − mw r 2 ⎜ i ⎟ = − mb vi 2 − mw vi 2 , so N =
2
2
⎝r⎠
1m v
2 b i
+ mw vi
1.8∆t
=
(7.5 m s)[ 12 (11 kg) + 1.3 kg]
1.8(4.5 s)
= 6.3 N .
13. Strategy Use conservation of energy. Let d = 2.05 m. Then, the ramp rises h = d sin 5.00°. The rotational inertia
of a uniform sphere is 52 mr 2 .
Solution Find the speed of the ball when it reaches the top of the ramp.
1
1
1
1
0 = ∆K + ∆U = mvf 2 + I ωf 2 − mvi 2 − I ωi 2 + mgh
2
2
2
2
2
2
v
1
1
2
1
1
⎛
⎞
⎛
⎞
⎛2
⎞⎛ v ⎞
= mvf 2 + ⎜ mr 2 ⎟ ⎜ f ⎟ − mvi 2 − ⎜ mr 2 ⎟ ⎜ i ⎟ + mgh
2
2⎝5
2
2⎝5
⎠⎝ r ⎠
⎠⎝ r ⎠
7
7
= mvf 2 − mvi 2 + mgh, so
10
10
vf = vi 2 −
d
h
5.00°
10
10
gh = (2.20 m s)2 − (9.80 m s 2 )(2.05 m) sin 5.00° = 1.53 m s .
7
7
14. Strategy Use conservation of angular momentum, Eq. (8-1), Eq. (8-14), and the relationship between period and
angular velocity.
Solution
(a) Since the angular momentum is conserved, the ratio is 1 .
(b) Since the rotational inertia is proportional to the square of the radius, ω =
Find the ratio of the angular velocities.
2
ωf ri 2 ⎛
1
⎞
8
= 2 =⎜
⎟ = 1.0 × 10
ωi rf
⎝ 1.0 × 10−4 ⎠
1 2 1⎛ L⎞ 2 1
I ω = ⎜ ⎟ ω = Lω.
2
2⎝ω ⎠
2
Find the ratio of the rotational kinetic energies.
K f ωf
=
= 1.0 × 108
Ki ωi
(c) The rotational kinetic energy is K rot =
(d) The period is related to the angular velocity by T =
2π
ω
.
Find the period of the star after collapse.
Tf ωi
ω
=
, so Tf = i Ti = (1.0 × 10−8 )(1.0 × 107 s) = 0.10 s .
Ti ωf
ωf
438
L
L
∝ 2.
I
r
Physics
Review and Synthesis: Chapters 6–8
15. Strategy Assume the collision time between the dart and the block is short so that the block’s motion during the
collision can be neglected. Let the dart be fired to the right and let the positive x-direction be to the right. Let the
origin be at the original position of the block. Use conservation of momentum during the collision and
conservation of energy after.
Solution Find the speed v of the dart and block just after the collision.
md
0.122 kg
pi = md vd = pf = (md + mb )v, so v =
vd =
(132 m s) = 3.144 m s.
0.122 kg + 5.00 kg
md + mb
Find the compression of the spring.
1
1
1
1
Wtotal = Wfriction + Wspring = − f k x − kx 2 = − µk Nx − kx 2 = − µk (md + mb ) gx − kx 2 = ∆K = 0 − (md + mb )v 2 ,
2
2
2
2
so 0 = kx 2 + 2µk (md + mb ) gx − (md + mb )v 2 . Solve for x.
x=
=
−2 µk (md + mb ) g ± [2 µk (md + mb ) g ]2 − 4k[−(md + mb )v 2 ]
2k
−(0.630)(5.122 kg)(9.80 m s 2 ) ± [(0.630)(5.122 kg)(9.80 m s 2 )]2 + (8.56 N m)(5.122 kg)(3.144 m s) 2
8.56 N m
= −3.69 m ± 4.42 m = 0.73 m or − 8.11 m
Since x > 0, the maximum compression is 0.73 m .
16. Strategy Use the conditions for equilibrium.
Solution Find the vertical components of the forces on each hinge.
mg (5.60 kg)(9.80 m s 2 )
ΣFy = 2 Fv − mg = 0, so Fv =
=
= 27.4 N .
2
2
Let the axis of rotation be a the midpoint of the left edge of the door. The only
horizontal forces are the horizontal components of the forces on the hinges,
therefore, these force are equal and opposite.
Στ = (0.735 m) Fh − (0.380 m)(5.60 kg)(9.80 m s 2 ) + (0.735 m) Fh = 0, so
(0.380 m)(5.60 kg)(9.80 m s 2 )
Fh =
= 14.2 N.
2(0.735 m)
The upper and lower horizontal forces on the hinges are 14.2 N
away from the door and 14.2 N toward the door, respectively.
439
0.760 m
Fv
0.735 m
Fh 0.380 m
2.030 m
0.735 m
mg
Fv
Fh
Review and Synthesis: Chapters 6–8
Physics
17. Strategy Use conservation of energy. The energy delivered to the fluid in the beaker plus the kinetic energies of
the pulley, spool, axle, paddles, and the block are equal to the work done by gravity on the block, which is
negative the change in the block’s gravitational potential energy. The rotational inertia of the pulley (uniform
solid disk) is 12 mp r 2 .
Solution Let the energy delivered to the fluid be E, the distance the block falls be h, and the rotational inertia of
the spool, axle, and paddles be Is = 0.00140 kg ⋅ m 2 . Since the radii of the pulley and the spool are the same (r),
their tangential speeds are the same, so let vp = vs = v.
2
2
1
1
1
1
1⎛1
1 ⎛v⎞
⎞⎛ v ⎞
mb vb 2 + I pωp 2 + I sωs 2 + E = mb vb 2 + ⎜ mp r 2 ⎟ ⎜ ⎟ + I s ⎜ ⎟ + E
2
2
2
2
2⎝2
2 ⎝r⎠
⎠⎝ r ⎠
The tangential speeds of the pulley and spool are equal to the speed of the block.
1
1
1 v2
1
1
1 v2
mb gh = mb vb 2 + mp v 2 + Is
+ E = mb v 2 + mp v 2 + Is
+ E , so
2
4
2 r2
2
4
2 r2
mb gh =
E = mb gh −
v 2 (2mb + mp + 2 I s r 2 )
4
= (0.870 kg)(9.80 m s 2 )(2.50 m) −
(3.00 m s)2 [2(0.870 kg) + 0.0600 kg + 2(0.00140 kg ⋅ m 2 ) (0.0300 m)2 ]
4
= 10.3 J .
18. Strategy Use conservation of linear momentum, the work-kinetic energy theorem, and Newton’s second law.
Solution According to Newton’s second law, the normal force of the ground on the players is N = (m1 + m2 ) g ,
where m1 = 85 kg and m2 = 95 kg. The force of friction is opposite the players direction of motion and has a
magnitude of f k = µk N = µk (m1 + m2 ) g . Find the initial speed v2 of the two-player combination.
pi = m1v1 = pf = (m1 + m2 )v2 , so v2 =
m1v1
.
m1 + m2
Find the distance d the players slide.
2
⎛ m1v1 ⎞
1
1
Wtotal = Wfriction = − f k d = − µk (m1 + m2 ) gd = ∆K = 0 − (m1 + m2 )v22 = − (m1 + m2 ) ⎜⎜
⎟⎟ , so
2
2
⎝ m1 + m2 ⎠
m12 v12
(85 kg) 2 (8.0 m s) 2
=
= 1.0 m .
d=
2
2 µk (m1 + m2 ) g 2(0.70)(85 kg + 95 kg)2 (9.80 m s 2 )
440
Physics
Review and Synthesis: Chapters 6–8
19. Strategy Since the collision is elastic, kinetic energy is conserved. Use conservation of linear momentum and
conservation of energy.
1
m v 2 = mA gh, so vAi = 2 gh , where h
2 A Ai
is the height fallen by bob A, 5.1 m. Since the mass of bob A is half that of bob B, let m = mA and 2m = mB .
Since kinetic energy is conserved, we have
1
1
1
1
mA gh = mA vAi 2 = mA vA 2 + mB vB2 = mvA 2 + mvB2 = mgh (1),
2
2
2
2
where vA and vB are the speeds of the bobs just after the collision.
Solution The kinetic energy of bob A just before is strikes bob B is
Use conservation of linear momentum to find vA in terms of vB .
pi = mvAi = pf = mvA + 2mvB , so vA = vAi − 2vB = 2 gh − 2vB . Substitute this into (1) and solve for vB .
1
1
2
m( 2 gh − 2vB )2 + mvB2 = m(2 gh − 4vB 2 gh + 4vB2 ) + mvB2 = mgh, so vB =
2 gh . Thus, we have,
2
2
3
vA = 2 gh − 2vB = 2 gh −
2 gh
4
2 gh = −
. Now, we use conservation of energy to find how high each bob
3
3
rises after the collision.
2
2 gh ⎞
1
1 ⎛
mgh
h 5.1 m
mghA = mvA 2 = m ⎜ −
, so hA = =
= 0.57 m .
⎟⎟ =
⎜
2
2 ⎝
3 ⎠
9
9
9
2
⎛ 2 2 gh ⎞
1
8mgh
4h 4(5.1 m)
2mghB = (2m)vB2 = m ⎜
, so hB =
=
= 2.3 m .
⎟⎟ =
⎜
2
9
9
9
⎝ 3 ⎠
441
Review and Synthesis: Chapters 6–8
Physics
20. Strategy Since the collision is elastic, kinetic energy is conserved. Use conservation of linear momentum. Let the
positive y-direction be along the shooter’s original velocity
Solution
(a) Let the mass of the marble be m, then the mass of the shooter is 3m. After the collision, let the speed of the
marble be v and the speed of the shooter be V.
1
1
1
(3m)Vi 2 = (3m)V 2 + mv 2 , so 3Vi 2 = 3Vx 2 + 3V y 2 + v 2 (1).
2
2
2
pix = 0 = pfx = 3mVx + mvx , so 0 = 3Vx − v sin 40° (2).
piy = 3mVi = pfy = 3mV y + mv y , so 3Vi = 3V y + v cos 40° (3).
3Vx
(4).
We have three equations and three unknowns (Vx , V y , and v). From (2), we have v =
sin 40°
Substituting this into (3) and solving for V y gives V y = Vi − Vx cot 40° (5). Substitute (4) and (5) into (1) and
solve for Vx .
⎛ 3Vx ⎞
3Vi 2 = 3Vx 2 + 3(Vi − Vx cot 40°)2 + ⎜
⎟
⎝ sin 40° ⎠
2
3Vi 2 = 3Vx 2 + 3(Vi 2 − 2ViVx cot 40° + Vx 2 cot 2 40°) +
0 = Vx 2 − 2ViVx cot 40° + Vx 2 cot 2 40° +
3Vx 2
9Vx 2
sin 2 40°
sin 2 40°
0 = Vx sin 40° − 2Vi cos 40° sin 40° + Vx cos 2 40° + 3Vx
2
0 = Vx (sin 2 40° + cos 2 40°) + 3Vx − 2Vi cos 40° sin 40°
0 = Vx (1) + 3Vx − 2Vi cos 40° sin 40°
4Vx = 2Vi cos 40° sin 40° = Vi sin 80°
V sin 80°
Vx = i
4
Use this result and (5) to find V.
2
⎤
⎛ V sin 80° ⎞ ⎡
⎛ Vi sin 80° ⎞
V = Vx 2 + V y 2 = Vx 2 + (Vi − Vx cot 40°)2 = ⎜ i
⎟ + ⎢Vi − ⎜
⎟ cot 40° ⎥
4
4
⎝
⎠ ⎣
⎝
⎠
⎦
2
= Vi
2
2
sin 2 80° ⎡ sin 80° cot 40° ⎤
sin 2 80° ⎡ sin 80° cot 40° ⎤
(3.2
m
s)
+ ⎢1 −
=
+ ⎢1 −
⎥
⎥ = 2.4 m s
16
4
16
4
⎣
⎦
⎣
⎦
(b) Substituting the result for Vx in (4) gives
3Vx
3V sin 80° 3(3.2 m s) sin 80°
v=
= i
=
= 3.7 m s .
sin 40°
4sin 40°
4sin 40°
(c) According to the way we set up the coordinate system, the tangent of θ is equal to Vx divided by V y instead
of the usual V y divided by Vx .
θ = tan −1
Vx
Vy
= tan −1
Vi sin 80° 4
sin 80°
= tan −1
= 19°
Vi − (Vi sin 80° 4) cot 40°
4 − sin 80° cot 40°
442
Physics
Review and Synthesis: Chapters 6–8
21. Strategy Use energy conservation to find the speed of Jones just before he grabs Smith. Then, use momentum
conservation to find the speed of both just after. Finally, again use energy conservation to find the final height.
Solution Find Jones’s speed, vJ .
1
mJ vJ 2 = mJ ghJ , so vJ = 2 ghJ .
2
Find the speed of both, v.
pi = mJ vJ = pf = (mJ + mS )v, so v =
m 2 ghJ
mJ vJ
= J
.
mJ + mS
mJ + mS
Find the final height, h.
2
⎛ m 2 ghJ ⎞
mJ 2 hJ
1
(78.0 kg) 2 (3.70 m)
⎟ , so h =
=
= 1.27 m .
(mJ + mS ) gh = (mJ + mS ) ⎜ J
⎜ m + mS ⎟
2
(mJ + mS )2 (78.0 kg + 55.0 kg)2
⎝ J
⎠
22. (a) Strategy Use the definition of angular acceleration.
Solution
∆ω 11 rad s − 0
α=
=
= 55 rad s 2 .
∆t
0.20 s
(b) Strategy Use Newton’s second law for rotation.
Solution Find the torque.
∆ω
11 rad s − 0
Στ = I α = I
= (1.5 kg ⋅ m 2 )
= 83 N ⋅ m .
∆t
0.20 s
(c) Strategy Use Eq. (5-21).
Solution Let ∆θ1 be the angle during spin-up and ∆θ 2 be the angle during spin-down.
ωf 2 − ωi 2 = ω 2 − 0 = 2α1∆θ1 , so ∆θ1 =
Find ∆θ1 + ∆θ 2 .
∆θ1 + ∆θ 2 =
ω2
−ω 2
. ωf 2 − ωi 2 = 0 − ω 2 = 2α 2 ∆θ 2 , so ∆θ 2 =
.
2α1
2α 2
⎞
ω 2 −ω 2 ω 2 ⎛ 1
1 ⎞ (11 rad s)2 ⎛
1
1
+
=
−
= 7.3 rad
⎜⎜
⎟
⎜⎜ −
⎟⎟ =
2
2α1 2α 2
2 ⎝ α1 α 2 ⎠
2
−9.8 rad s 2 ⎟⎠
⎝ 55 rad s
(d) Strategy Use Eq. (5-18) and the relationship between angular speed and linear speed.
Solution Find the speed of a point halfway along the radius of the disk 0.20 s after the accelerating torque is
removed.
v
ωf − ωi = − ωi = α∆t , so
r
0.115 m ⎡
v = r (α∆t + ωi ) =
(−9.8 rad s 2 )(0.20 s) + (11 rad s) ⎤ = 0.52 m s .
⎣
⎦
2
443
Review and Synthesis: Chapters 6–8
Physics
23. Strategy Use the work-kinetic energy theorem and Newton’s second law.
Solution ΣFy = N − mg cos θ = 0, so N = mg cos θ . Thus, f k = µk mg cos θ .
1
Wtotal = Wgrav + Wfriction = mgd sin θ − µk mg cos θ d = ∆K = mvf 2 , so
2
vf = 2 gd (sin θ − µk cos θ ) = 2(9.80 m
s 2 )(0.300
fk
N
y
θ
m)(sin 60.0° − 0.38cos 60.0°)
x d
= 2.0 m s .
mg cos θ
mg sin θ
mg
θ
24. Strategy The cylinder falls a vertical distance h = d sin θ = (0.300 m) sin 60.0° as it rolls down the incline. The
rotational inertia of a uniform solid cylinder is 12 mr 2 . Use conservation of energy.
Solution Find the cylinder’s final speed.
2
1
1
1
1⎛1
⎞⎛ v ⎞
mv 2 + I ωf 2 − 0 + 0 − mgh = mvf 2 + ⎜ mr 2 ⎟ ⎜ f ⎟ − mgd sin θ
2 f
2
2
2⎝2
⎠⎝ r ⎠
3
4
4
gd sin θ =
= mvf 2 − mgd sin θ , so vf =
(9.80 m s 2 )(0.300 m) sin 60.0° = 1.84 m s .
4
3
3
0 = ∆K + ∆U =
d
θ
25. Strategy Use conservation of linear momentum.
Solution
y
mb vb
.
pix = mb vb = pfx = (mb + mc )vfx , so vfx =
mb + mc
N
mc vc
.
piy = mc vc = pfy = (mb + mc )vfy , so vfy =
mb + mc
vc
vb
Compute the magnitude of the final velocity.
⎛ mb vb
v = vfx 2 + vfy 2 = ⎜⎜
⎝ mb + mc
2
⎞ ⎛ mc vc
⎟⎟ + ⎜⎜
⎠ ⎝ mb + mc
2
(mb vb )2 + (mc vc )2
⎞
=
⎟⎟
mb + mc
⎠
[(2.00 kg)(2.70 m s)]2 + [(1.50 kg)(−3.20 m s)]2
= 2.06 m s
2.00 kg + 1.50 kg
Compute the angle.
=
θ=
vfy
tan −1
vfx
=
mcvc
mb + mc
−
1
tan
mb vb
mb + mc
= tan −1
mc vc
mb vb
= tan −1
(1.50 kg)(−3.20 m s)
= − 41.6°
(2.00 kg)(2.70 m s)
The velocity of the block and the clay after the collision is 2.06 m s at 41.6° S of E .
444
x
vf
d sin θ
Physics
Review and Synthesis: Chapters 6–8
26. (a) Strategy Use conservation of angular momentum.
Solution Find the tangential speed of the skaters after they grab the rods.
v
v
Li = I1iω1i + I 2iω2i = m1r 2 1 + m2 r 2 2 = m1rv1 + m2 rv2 = Lf = I f ωf = m1rv + m2 rv, so
r
r
m1v1 + m2 v2 = (m1 + m2 )v.
Solve for the tangential speed.
m v + m2 v2 (60.0 kg)(6.0 m s) + (30.0 kg)(2.0 m s)
=
= 4.7 m s .
(m1 + m2 )v = m1v1 + m2 v2 , so v = 1 1
60.0 kg + 30.0 kg
m1 + m2
(b) Strategy and Solution According to the RHR, the angular momentum is upward, away from the ice before
and after the collision. Angular momentum is conserved in magnitude and direction.
27. (a) Strategy Use conservation of angular momentum, since no external torques act on the two-disk system.
Solution Find the final angular velocity.
ωi
ωi
Iω
Iiωi
=
=
.
Lf = I f ωf = Li = I iωi , so ωf = i i =
2
2
2
2
1
If
1 + mr [2(MR 2)] 1 + mr (MR 2 )
Ii + 2 mr
(b) Strategy and Solution
The total angular momentum does not change, since no external torques act on the system.
(c) Strategy Compute the initial and final total kinetic energies and compare their values.
Solution
1
1⎛1
1
⎞
Ki = Iiωi 2 = ⎜ MR 2 ⎟ ωi 2 = MR 2ωi 2
2
2⎝2
4
⎠
⎛
1
1⎛1
1
ωi
⎞
K f = I f ωf 2 = ⎜ MR 2 + mr 2 ⎟ ⎜⎜
2
2
2⎝2
2
⎠ ⎜ 1 + mr 2
MR
⎝
=
(
)
2
1 1 MR 2 ω 2
i
2 2
1 MR 2 + 1 mr 2
2
2
2
2
2
⎞
1 MR 2ω
⎛
⎞
i
⎟ = 1 ⎛ 1 MR 2 + 1 mr 2 ⎞ ⎜
2
⎟
⎟⎜ 1
⎟
2 ⎜⎝ 2
2
⎠ ⎝ 2 MR 2 + 12 mr 2 ⎟⎠
⎟
⎠
2
MR ωi
Ki
= 4
=
2
2
mr
1+ 2
1 + mr 2
1
MR
MR
So, K f ≠ Ki . Therefore, the answer is yes; the kinetic energy changes.
28. (a) Strategy The candy is release with a horizontal speed equal to the tangential speed of the pocket of the
rotating wheel. Use the relationship between angular and tangential speed and the equations of motion for a
changing velocity.
Solution Find the time it takes for the candy to land.
1
1
2 ∆y
∆y = a y (∆t )2 = (− g )(∆t )2 , so ∆t = −
.
2
2
g
Find the candy’s distance from its starting point.
⎛ 2π rad ⎞
2∆y
2(−0.240 m)
v = r ω , so ∆x = v∆t = r ω −
= (0.120 m)(1.60 Hz) ⎜
= 0.267 m .
⎟ −
g
9.80 m s 2
⎝ cycle ⎠
(b) Strategy Use the relationship between radial acceleration and angular speed.
Solution Find the radial acceleration of the candy.
ar = ω 2 r = (1.60 Hz)2 (2π rad cycle)2 (0.120 m) = 12.1 m s 2
445
Review and Synthesis: Chapters 6–8
Physics
29. (a) Strategy Consider the work-kinetic energy theorem and the impulse momentum theorem.
Solution Since the Romulan ship is twice as massive as the Vulcan ship, the Romulan ship will not travel as
far as the Vulcan ship for the same engine force, since ∆x = (1 2)a (∆t ) 2 = (1 2)( F m)(∆t ) 2 . Since
W = F ∆x = ∆K ,
the Vulcan ship will have the greater kinetic energy . Since ∆p = F ∆t ,
the ships will have the same momentum .
(b) Strategy Consider the work-kinetic energy theorem and the impulse momentum theorem.
Solution Since the distances and the forces are the same, and since W = F ∆x = ∆K ,
the ships will have the same kinetic energy . Since ∆x = (1 2)a (∆t ) 2 = (1 2)( F m)(∆t ) 2 , the more massive
Romulan ship will have to fire its engines longer than the Vulcan ship to travel the same distance. Since
∆p = F ∆t and the forces are the same, the Romulan ship will have the greater momentum .
(c) Strategy Refer to parts (a) and (b).
Solution For part (a), we have the following:
Vulcan:
⎡ F
⎤ (9.5 × 106 N) 2 (100 s) 2
∆K = W = F ∆ x = F ⎢
= 6.9 × 1012 J
( ∆t ) 2 ⎥ =
2(65, 000 kg)
⎣ 2m
⎦
∆p = F ∆t = (9.5 × 106 N)(100 s) = 9.5 × 108 kg ⋅ m s
Romulan:
⎡ F
⎤ (9.5 × 106 N) 2 (100 s) 2
∆K = W = F ∆ x = F ⎢
= 3.5 × 1012 J
( ∆t ) 2 ⎥ =
2
m
2(130,
000
kg)
⎣
⎦
∆p = F ∆t = (9.5 × 106 N)(100 s) = 9.5 × 108 kg ⋅ m s
In part (a), the momenta are the same, 9.5 × 108 kg ⋅ m s , but the kinetic energies differ:
Vulcan at 6.9 × 1012 J and Romulan at 3.5 × 1012 J.
For part (b), we have the following:
Vulcan:
∆K = W = F ∆x = (9.5 × 106 N)(100 m) = 9.5 × 108 J
Since K =
1 2 p2
mv =
, p = 2mK = 2(65, 000 kg)(9.5 × 108 J) = 1.1× 107 kg ⋅ m s.
2
2m
Romulan:
∆K = W = F ∆x = (9.5 × 106 N)(100 m) = 9.5 × 108 J
p = 2mK = 2(2 × 65, 000 kg)(9.5 × 108 J) = 1.6 × 107 kg ⋅ m s.
In part (b), the kinetic energies are the same, 9.5 × 108 J, but the momenta differ:
Vulcan at 1.1× 107 kg ⋅ m s and Romulan at 1.6 × 107 kg ⋅ m s.
446
Physics
Review and Synthesis: Chapters 6–8
30. (a) Strategy Use conservation of energy.
Solution Let d be the distance moved along the incline by m2 . Both masses move the same distance and
have the same speed, since they are connected by a rope.
1
1
0 = ∆K + ∆U = m1v 2 + m2 v 2 − 0 − 0 + m1 g (− d sin θ ) + m2 gd sin φ , so
2
2
v=
2 gd (m1 sin θ − m2 sin φ )
=
m1 + m2
2(9.80 m s 2 )(2.00 m)[(6.00 kg) sin 36.9° − (4.00 kg) sin 45.0°]
6.00 kg + 4.00 kg
= 1.7 m s .
(b) Strategy Use Newton’s second law.
Solution Let the positive direction be along the inclines from left to right.
For m1: ΣF = −T + m1 g sin θ = m1a, so T = m1 g sin θ − m1a.
For m2 : ΣF = T − m2 g sin φ = m2 a, so T = m2 g sin φ + m2 a.
Solve for the acceleration.
m2 g sin φ + m2 a = m1 g sin θ − m1a, so
g (m1 sin θ − m2 sin φ ) (9.80 m s 2 )[(6.00 kg) sin 36.9° − (4.00 kg) sin 45.0°]
=
= 0.76 m s 2 .
a=
6.00 kg + 4.00 kg
m1 + m2
Find the speed.
vf 2 − vi 2 = v 2 − 0 = 2ad , so v = 2ad = 2(0.76 m s 2 )(2.00 m) = 1.7 m s .
31. Strategy Refer to the figure and use conservation of energy.
Solution
(a) According to the graph, the particle’s potential energy is −550 J . Since E = K + U , the kinetic energy of
the particle is K = E − U = −100 J − (−550 J) = 450 J .
(b) The total energy is as given, −100 J . According to the graph, the potential energy is −100 J . The
kinetic energy is K = E − U = −100 J − (−100 J) = 0 .
(c) The kinetic energy is K = E − U = −100 J − (−300 J) = 200 J .
(d)
The particle has a kinetic energy of 450 J at t = 0, and we are told the motion is to the left. The particle
will continue moving left but the kinetic energy will decrease by 450/4.5 J for every cm of travel until it
reaches x = 1 cm. At this point K = 0, and the particle has stopped instantaneously. It will next move to
the right with an increasing K until it reaches x = 5.5 cm. At this point K = 450 J, and this kinetic energy
will be maintained as it continues moving right until it reaches x = 11 cm. At this point, its kinetic energy
will decrease by 450/2.5 J for every cm of travel until it reaches x = 13.5 cm. At this point K = 0, and
the particle has again stopped instantaneously. It will then turn around again.
447
Review and Synthesis: Chapters 6–8
Physics
32. (a) Strategy Use conservation of angular momentum at the moment of impact.
Solution Li = I ωi = Lf = I ωf + rp, where I is the rotational inertia of the blade, r is the distance from the
center of the blade to the location of impact, and p = mv is the momentum of the stone just after it is struck.
Find the speed of the stone.
I ωi = I ωf + rp = I ωf + rmvtan
I (ωi − ωf ) = rmvtan
2
2
1
1
I (ωi − ωf ) 12 ML (ωi − ωf ) 12 M (2r ) (ωi − ωf ) Mr (ωi − ωf )
=
=
=
rm
rm
rm
3m
(2.0 kg)(0.25 m)[2π (60 rev s − 55 rev s)]
=
= 52 m s
3(0.10 kg)
vtan =
(b) Strategy Find time it takes for the stone to reach the house. Then use this time to find the distance the stone
falls just before it reaches the window.
Solution Find the time.
∆x
10.0 m
∆x = v∆t , so ∆t =
=
= 0.191 s.
vtan 52.4 m s
Find the distance the stone falls.
1
1
∆y = − g (∆t )2 = − (9.8 m s 2 )(0.191 s)2 = − 0.18 m
2
2
Since 0.18 m is less than half of 1.00 m (0.50 m), the stone hits the window.
33. Strategy Use conservation of energy. m is the mass of one wheel. M is the total mass of the system. v is the speed
of the center of mass of the system (which is the same as the speed of a point on either wheel).
Solution
1 2 1 2
1
1
I ω = mv = K trans for one wheel. K rot,total = 2 ⋅ mv 2 = mv 2 and K trans,total = Mv 2.
2
2
2
2
K total = U i
1
mv 2 + Mv 2 = MgH
2
v 2 (2m + M ) = 2MgH
(a) K rot =
v=
2MgH
=
2m + M
2(80.0 kg)(9.8 m s 2 )(20.0 m)
= 19 m s
2(1.5 kg) + 80.0 kg
(b) Since the speed depends upon the combined total mass of the system, the speed at the bottom would not be
the same for a less massive rider. The answer is no .
448
Physics
Review and Synthesis: Chapters 6–8
34. Strategy Find the change is height from the initial height to the height at which the vine breaks; then use the
change in height to find Tarzan’s speed when the vine breaks. Use Newton’s second law to find the tension when
the vine breaks.
Solution
(a) Find L, the length of the vine.
opposite
5.00 m
5.00 m 5.00 m
sin θi =
, so L =
=
=
= 5.77 m.
hypotenuse
L
sin θi
sin 60°
Find the change in height.
5.00 m
(cos θ f − cos θi )
∆y = yf − yi = L cos θf − L cos θi =
sin θi
Find Tarzan’s speed when the vine breaks.
5.00 m
(cos θ f − cos θi )
vf2 − vi2 = v 2 − 0 = 2 g ∆y = 2 gL(cos θ f − cos θi ) = 2 g
sin θi
5.00 m
5.00 m
v = 2g
(cos θ f − cos θi ) = 2(9.80 m s 2 )
(cos 20.0° − cos 60.0°) = 7.05 m s
sin θi
sin 60.0°
Find the tension.
mv 2
ΣF = T − mg cos θf =
, so
r
⎛ v2
⎞
mv 2
mgv 2
T=
+ mg cos θf =
+ mg cos θ f = mg ⎜
+ cos θ f ⎟
⎜
⎟
r
gL
⎝ gL
⎠
⎡
⎤
(7.05 m s)2
= (900.0 N) ⎢
+ cos 20°⎥ = 1.64 kN
⎢⎣ (9.80 m s 2 )(5.77 m)
⎥⎦
(b) At the moment of the vine breaking, the distance to ground level is
8.00 m − L cos θ f = 8.00 m − (5.77 m) cos 20.0° = 2.58 m;
and the distance to the river’s edge is
L sin θf = (5.77 m) sin 20.0° = 1.97 m.
The time it takes for Tarzan to reach ground level is given by
1
1
∆y = −v sin θf ∆t − g (∆t ) 2 , or 0 = g (∆t )2 + v sin θf ∆t + ∆y.
2
2
Using the quadratic formula, we find ∆t = 0.52 s. The horizontal distance traveled in this time is
∆x = v cos θ f ∆t = (7.05 m s) cos 20.0°(0.52 s) = 3.4 m. Since 3.4 m > 1.97 m, Tarzan lands safely on the
other side. The answer is yes .
449
Review and Synthesis: Chapters 6–8
Physics
35. Strategy Use the relationship between energy and work to find the boy’s speed just before his friend lands on the
sled and the speed when the two reach the bottom. Use conservation of momentum to find the speed of the boys
just after the friend lands on the sled.
Solution Find the speed of the boy.
1
∆K = m1v12 − 0 = −∆U + Wfriction = m1gh1 − fd1 = m1gd1 sin θ − µ m1g cos θ d1, so
2
v1 = 2 gd1(sin θ − µ cos θ ) = 2(9.8 m s 2 )(20 m)(sin15° − 0.12 cos15°) = 7.48 m s.
Find the initial speed of the two boys.
(60 kg)(7.48 m s)
m1v1
pi = m1v1 = pf = (m1 + m2 )v2 , so v2 =
=
= 4.08 m s.
60 kg + 50 kg
m1 + m2
Find the final speed of the two boys.
∆K = −∆U + Wfriction
1
1
(m1 + m2 )v32 − (m1 + m2 )v22 = (m1 + m2 ) gh2 − fd 2 = (m1 + m2 ) gd 2 sin θ − µ (m1 + m2 ) g cos θ d 2
2
2
v32 = 2 gd 2 sin θ − 2µ g cos θ d 2 + v22
v3 = 2 gd 2 (sin θ − µ cos θ ) + v22
= 2(9.8 m s 2 )(50 m)(sin15° − 0.12 cos15°) + (4.08 m s)2 = 13 m s
36. Strategy Use conservation of momentum and the equations for motion with a constant acceleration.
Solution
(a) The banana will fall at the same rate as the monkey; therefore, you should throw the banana directly at the
monkey.
3.33 m + 1.67 m
5.00
tan θ =
, so θ = tan −1
= 59.0° above the horizontal .
3.00 m
3.00
(b) Since the banana will fall at the same rate as the monkey, regardless of the launch speed of the banana, the
launch angle is the same for all launch speeds. Relatively high launch speeds will reach the monkey relatively
sooner (and higher); relatively low launch speeds will reach the monkey relatively later (and lower).
(c) Find the time it takes the banana to reach the monkey.
1
2∆y
2(−1.67 m)
∆y = − g (∆t ) 2 , so ∆t = −
= −
= 0.5838 s = t.
2
g
9.80 m s 2
The banana reaches the monkey when it has traveled 3.00 m.
∆x
3.00 m
∆x = vx ∆t = v cos θ∆t , so v =
=
= 9.98 m s .
∆t cos θ (0.5838 s) cos 59.0°
450
Physics
Review and Synthesis: Chapters 6–8
(d) The speed of the monkey just before the collision is given by vmy = − gt and vmx = 0. The speed of the banana
at this time is given by vby = v sin θ − gt and vbx = v cos θ . Use conservation of momentum.
mvbx = (m + M )vfx , so
m
m
0.20 kg
vfx =
vbx =
v cos θ =
(9.98 m s) cos 59.0° = 0.321 m s.
m+M
m+M
0.20 kg + 3.00 kg
mvby + Mvmy = (m + M )vfy , so
vfy =
mvby + Mvmy
=
m(v sin θ − gt ) + M (− gt )
m
=
v sin θ − gt
m+M
m+M
m+M
0.20 kg
=
(9.98 m s) sin 59.0° − (9.80 m s 2 )(0.5835 s) = −5.18 m s.
0.20 kg + 3.00 kg
The time it takes for the monkey to hit the ground is given by
1
1
∆y = vfyt2 − gt22 , or 0 = gt22 − vfyt2 + ∆y = (4.90 m s 2 )t22 + (5.18 m s)t2 − 5.33 m.
2
2
Using the quadratic formula, we find t = 0.64 s. The horizontal distance is
d = vfxt = (0.321 m s)(0.64 s) = 0.21 m .
MCAT Review
1. Strategy Use conservation of momentum.
Solution
pi = mvi = pf = mvf + pwall , so pwall = m(vi − vf ) = (0.2 kg)[2.0 m s − (−1.0 m s)] = 0.6 kg ⋅ m s.
The correct answer is D .
2. Strategy Use Hooke’s law.
Solution Let up be the positive direction. The gravitational force on the mass is
F = mg = (0.10 kg)(−9.80 m s 2 ) = −0.98 N. Solving for the spring constant in Hooke’s law, we have
k =−
−0.98 N
F
=−
= 6.5 N m. Thus, the correct answer is D .
x
0.15 m
3. Strategy The net torque is zero.
Solution
m) − (1.0 × 10−7
0.60 m
s 2 )(0.40
Στ = 0 = F (0.60
kg)(9.80 m
m), so
7
2
−
(1.0 × 10 kg)(9.80 m s )(0.40 m)
F=
= 6.5 × 10−7 N.
0.60 m
The correct answer is B .
451
F
0.40 m
mg
Review and Synthesis: Chapters 6–8
Physics
4. Strategy Determine the speed of the first ball just before in collides with the second. The collision is completely
inelastic; that is, the balls stick together. Use conservation of momentum to find the speed of the balls after the
collision.
Solution Find the speed of the first ball just before the collision.
vfx − vix = v1 − 0 = a x ∆t , so v1 = (10 m s 2 )(2.0 s) = 20 m s.
Find the speed v of the balls just after the collision.
m1v1
(0.50 kg)(20 m s)
pi = m1v1 = pf = (m1 + m2 )v, so v =
=
= 6.7 m s.
m1 + m2
0.50 kg + 1.0 kg
The correct answer is B .
5. Strategy Use Newton’s second law and Eq. (6-27).
Solution The gravitational force working against the motion of the car as it climbs the hill is
mg sin10°, so the additional power required is
10°
mg
Pcar = − Pgrav = − Fv cos180° = (mg sin10°)v = (1000 kg)(10 m s 2 ) sin10°(15 m s)
= 1.5 × 105 × sin10°
mg sin 10°
W.
The correct answer is D .
6. Strategy Find the vertical distance the patient would have climbed had the treadmill been stationary (and very
long). Then, find the work done by the patient on the treadmill.
Solution The “distance” walked along the incline is (2 m s)(600 s) = 1200 m.
Thus, the vertical distance climbed is (1200 m) sin 30° = 600 m. The work done is
W = Fd = mgd = (90 kg)(10 m s 2 )(600 m) = 0.54 MJ.
0m
120
600 m
30°
The correct answer is C .
7. Strategy Find the angle between the force exerted by the patient and the patient’s velocity. Use Eq. (6-27).
Solution The force due to gravity is down, so the force exerted by the patient is up. The velocity is
directed at the angle of the incline, or 30° above the horizontal, so the angle between the force and
the velocity is 60°. Compute the mechanical power output of the patient.
P = Fv cos θ = mgv cos θ = (100 kg)(10 m
s 2 )(3
F
60°
v
m s) cos 60° = 1500 W
The correct answer is B .
8. Strategy and Solution The force pushing each friction pad is normal to the wheel; that is, it is the normal force
in f k = µk N . Solve for the normal force.
f
20 N
N= k =
= 50 N
0.4
µk
This is the total force. The force pushing each friction pad is half this, or 25 N. The correct answer is B .
452
Physics
Review and Synthesis: Chapters 6–8
9. Strategy Find the average tangential speed at the friction pads. Then, use the relationship between tangential
speed and radial acceleration.
Solution
The average tangential speed is v =
ar =
4800 m 1 min
×
= 4.0 m s. The radial acceleration is
20 min 60 s
v 2 (4.0 m s) 2
=
= 50 m s 2 . The correct answer is D .
r
0.3 m
10. Strategy Use the work-kinetic energy theorem.
Solution The work done by friction on the wheel is W = − f k d , where d is the linear distance the wheel passes
between the pads before it stops. Relate d to the kinetic energy of the wheel.
K
Wtotal = − f k d = ∆K = 0 − Ki , so d = i .
fk
Divide d by the circumference of a circle with radius 0.3 m to find the number of rotations.
Ki
d
30 J
=
=
= 0.8 rotations
2π r 2π rf k 2π (0.3 m)(20 N)
Since 0.8 < 1, the correct answer is A .
11. Strategy Compute the average mechanical power output of the cyclist and compare it to the power consumed by
the wheel at the friction pads.
Solution The metabolic power available for work is 535 W − 85 W = 450 W. Since the efficiency is 20%, the
average mechanical power output of the cyclist is 0.20 × 450 W = 90 W. The average tangential speed of the
4800 m 1 min
×
= 4.0 m s. Therefore, the power consumed by the friction pads is
wheel is v =
20 min 60 s
P = f k v = (20 N)(4.0 m s) = 80 W. Thus, the difference between the average mechanical power output of the
cyclist and the power consumed by the wheel at the friction pads is 90 W − 80 W = 10 W.
The correct answer is B .
12. Strategy and Solution Increasing the force on the friction pads would increase the power consumed by the wheel
at the friction pads (because P = Fv). So, if the cyclist is pedaling at the same rate and the power consumed by
the friction pads increases, the difference between the two decreases and the fraction of mechanical power output
of the cyclist consumed by the wheel at the friction pad increases. Thus, the correct answer is D .
13. Strategy Relate the cyclist’s average metabolic rate to the energy released per volume of oxygen consumed, the
time on the bike, and volume of oxygen consumed.
Solution The cyclist’s average metabolic rate while riding is 535 W. The total energy used during 20 minutes is
60 s
(535 W)(20 min)
= 642, 000 J. The total energy released by the consumption of oxygen is (20, 000 J L)V ,
1 min
where V is the volume of oxygen consumed. Equating these two expressions and solving for V gives the number
of liters of oxygen the cyclist consumes.
642, 000 J
(20, 000 J L)V = 642, 000 J, so V =
= 32 L ≈ 30 L. The correct answer is B .
20, 000 J L
453
Review and Synthesis: Chapters 6–8
Physics
14. Strategy and Solution Since the force has been reduced by 50% and the distance has been doubled, the cyclist
does the same amount of work [W = 0.50 F (2∆x) = F ∆x]. So, the energy transmitted in the second workout is
equal to the energy transmitted in the first. The correct answer is C .
15. Strategy The circumference of a circle is C = 2π r. A wheel moves a distance equal to its circumference during
each rotation. The wheel rotates twice during each rotation of the pedals.
Solution The circumference of a circle with a radius of 0.15 m is 2π (0.15 m). The circumference of a circle with
a radius of 0.3 m is 2π (0.3 m). During each rotation of the pedals, a point on the wheel at a radius of 0.3 m
moves a distance 2[2π (0.3 m)]. The ratio of the distance moved by a pedal to the distance moved by a point on
the wheel located at a radius of 0.3 m in the same amount of time is
2π (0.15 m)
= 0.25.
2[2π (0.3 m)]
The correct answer is A .
16. Strategy Use the definition of power.
Solution
∆E
∆E ⎛ 300 kcal ⎞ ⎛ 4186 J ⎞ ⎛ 1 min ⎞
P=
, so ∆t =
=⎜
⎟⎜
⎟⎜
⎟ = 41.9 min. The correct answer is D .
P ⎝ 500 W ⎠ ⎝ 1 kcal ⎠ ⎝ 60 s ⎠
∆t
17. Strategy Consider the distance a point on the wheel travels for each situation.
Solution The circumference of a circle with a radius of 0.3 m is 2π (0.3 m). The
circumference of a circle with a radius of 0.4 m is 2π (0.4 m). During each rotation, a point
on a wheel travels a distance equal to the circumference. The force on the wheel is the same
in each case, but the distance traveled by a point on the wheel is greater for a greater radius.
In this case, the distance is 0.4 m (0.3 m) = 1.33 times farther or 33%. Since work is equal
to the product of force times distance, the work done on the wheel per revolution is 33%
more. Thus, the correct answer is C .
454
0.3 m
0.4 m