Practice Problems
1.
D
Net force equals Fc, which is toward the center of the
circle.
2.
B
Turn to the left results in the right side of the bus being
in "front" of you as you continue forward due to inertia.
3.
B
Centripetal force generated by friction between road
and tires is not sufficient for the turn at high speed.
4.
B
Without the tension in the string to supply the
centripetal force, the ball continues in a straight line.
5.
B
Fc = mv2/r = m(2r/T)2/r = 42mr/T2, since 42m/T2 is
constant  Fc  r.
6.
C
At the bottom of the vertical circle:
Fnet = Fc = Fn – Fg > 0 (upward)  Fn > Fg.
7.
A
At the top of the vertical circle, both Ft and Fg are
downward, which is the direction of Fc  Fc = Ft + Fg.
8.
C
Newton's third law states that for every force there is
an equal but opposite force  the pulls are equal.
d.
tan = Ft-y/Ft-x = Fg/Fc
tan = 20/50   = 22o
14. a.
Ft = Fc – Fg = mv2/r – mg
Ft = (2 kg)(5 m/s)2/(1 m) – (2 kg)(10 m/s2) = 30 N
b.
Ft = Fc + Fg = mv2/r + mg
Ft = (2 kg)(5 m/s)2/(1 m) + (2 kg)(10 m/s2) = 70 N
15. a.
Ft = Fg + Fc = mg + mv2/r
15 N = (1kg)(10 m/s2) +(1 kg)v2/(2 m) v = 3.16 m/s
b.
Ug + K = Ug' + K' mgh = ½mv2 
(10 m/s2)h = ½(3.16 m/s)2 h = 0.50 m
16. a.
Between Earth and Moon
On the Earth's surface
the same
greater
b.
Between Earth and Moon
On the Earth's surface
less
less
17.
Mass
Radius (x Earth)
Acceleration (x gEarth)
m = mEarth
r = rEarth
g
9.
A
The net force is Fg, which provides the Fc necessary to
keep the astronaut in orbit.
10.
v v = 2r/T = 2(50 m)/16 s = 20 m/s
direction of v
direction of ac
11. a.
b.
v = 2r/T = 2(1.5 x 1011 m)/3.2 x 107 s = 3.0 x 104 m/s
c.
aC = v2/r = (3.0 x 104 m/s)2/1.5 x 1011 m = 6 x 10-3 m/s2
12. a.
Fc = Ff  mv2/r = mg
v2/50 m = (0.8)(10 m/s2)  v = 20 m/s
b. (1)
19.
m = mEarth
r = ½rEarth
4g
m = 2mEarth
r = rEarth
2g
m = ½mEarth
r = rEarth
½g
GmMm/rM2 = GmEm/rE2
rE2/rM2 = mE/mM = 100  rM/rE = 10
20.
Which changes would decrease the acceleration
II and III
due to gravity on the earth's surface?
Which changes would increase the acceleration
I and IV
due to gravity on the earth's surface?
Which changes would decrease the acceleration
II only
due to gravity on the moon?
Which changes would increase the acceleration
I only
due to gravity on the moon?
21. a.
g = GM/r2  gMoonrMoon2/mMoon = gEarthrEarth2/mEarth
gMoon(1)2/1 = g(42)/81 gMoon = 16g/81 = 0.2 g
b.
Mass doesn't change: 50 kg
Yes, because mass cancels out of the equation.
c.
(2)
Fg = mg = (50 kg)(0.2x 10 m/s2) = 100 N
No, because the coefficient of friction is less.
13. a.
Ft-x = Fc = mv2/r
Ft-x = (2 kg)(5 m/s)2/(1 m) = 50 N
b.
Ft-y = Fg = mg
Ft-y = (2 kg)(10 m/s2) = 20 N
c.
Ft = (Ft-x2 + Ft-y2)½
Ft = (502 + 202)½ = 54 N
¼g
g = GM/r2
g = (6.67x10-11N•m2/kg2)(6.4x1023kg)/(3.4x106m)2=3.7m/s2
south
T = 365 days x 24 hr/day x 3600 s/hr = 3.2 x 107 s
r = 2rEarth
18.
east
ac ac = v2/r = (20 m/s)2/(50 m) = 8 m/s2
m = mEarth
22.
B
Torque is the greatest when rsin is maximum, which is
when the distance is the greatest and the angle is 90o.
23.
A
Sliding is greater (Fgsin = ma) than rolling
(Fgsin = (1 +  )ma), and s doesn't effect acceleration.
24.
A
Most of the mass is along its rim   in I = mr2 is the
greatest.
25.
D
26.
C
27.
B
Fr = ma: the same force would generate a greater
acceleration for the object with the smallest .
Placing all the rod's mass at its center 
rbmbg = rrmrg  (0.25 m)(1 kg) = (0.25 m)(mrod)  mr = 1 kg
37. a.
Ug = Krollingand h = 1.0 m(sin) = 0.5 m
mgh = ½(1 +  )mv2 v = [g/(1 +  )]½
b.
Hoop ( = 1)
v = [g/(1 +  )]½ = [10/(1 + 1)]½ = 2.24 m/s
Cylinder ( = 1/2) v = [g/(1 +  )]½ = [10/(1 + 1/2)]½ = 2.58 m/s
(1 m)B = (3 m)(1 kg)  B = 3 kg
(1 m)(1 kg + 3 kg) = (2 m)(A)  A = 2 kg  1 + 3 + 2 = 6
Sphere ( = 2/5)
v = [g/(1 +  )]½ = [10/(1 + 2/5)]½ = 2.67 m/s
Box ( = 0)
v = [g/(1 +  )]½ = [10/(1 + 0)]½ = 3.16 m/s
28.
The blocks to right of the table's edge are not balanced
by an equal number of blocks to the left.
29. a.
A
1 = rF1 = (1 m)(45 N) = 45 m•N
b.
1 = rF2
45 m•N = (0.4 m)F2  F2 = 113 N
30. a.
Fg-|| = Fgsin= (5 kg)(10 m/s2)sin30 = 25 N
b.
Fg-|| = Frolling = (1 +  )ma
25 N = (1 + ½)(5 kg)a  a = 3.3 m/s2
31. a.
Ff = Fn = (0.80)(250 N) = 200 N
b.
Ff = Fc = mv2/r
200 N = (25 kg)v2/(2 m)  v = 4 m/s
c.
Fr = ma
50 N = ½(200 kg)a  a = 0.5 m/s2
d.
vt = vo + at
4 m/s = 0 + (0.5 m/s2)t  t = 8 s
e.
Increase
32.
child = plank
(x m)(250 N) = (0.5 m)(750 N)  x = 1.5 m
33.
F1
F11500 + 15,000
(20)F1 = (10)(15,000) + (5)(150,000)  F1 = 45,000 N
F1 + F2 = 1500g + 15,000g
F2
45,000 N + F2 = 15,000 N + 150,000 N  F2 = 120,000 N
34. a.
mstudent = 35.1 kg + 31.6 kg
mstudent = 66.7 kg
b.
rcm = (r1m1 + r2m2)/(m1 + m2)
rcm = [(172)(35.1) + (0)(31.6)]/(66.7) = 90.5 cm
35.
A = cm
A
(8 m)FA = (5.5 m)(22000 N)  FA = 15,000 N
FA + FB = Fg
B
15,000 N + FB = 22,000 N  FB = 7,000 N
36.
cc = c
(5 m)Fguy wiresin37 = (5 m)(200 N)sin53  F = 265 N
c.
Box-sphere-Cylinder-Hoop
38.
Ug = K'rolling  mgh = ½(1 + ½)mv2
(10 m/s2)(0.50 m) = ¾v2  v = 2.6 m/s
39. a.
Fg = Fc
mg = mv2/r  v = (rg)½ = 10 m/s
b.
Ug-A = U'g-B + K'rolling-B 
mgH + 0 = mgh + ½(1 +  )mv2
(10)H = (10)(20) + ½( 1 +2/5)(10)2 H = 27 m
40.
Ub = K'b + K'r-p
mbgh = ½mbv2 + ½mpv2 = ½(mb + mp)v2
(1 kg)(10 m/s2)(1 m) = ½(1.0 kg + 1.0 kg)v2 v = 3.2 m/s
41.
Um2 + Um1 = U'm2 + U'm1 + K'r-M + K'm1 + K'm2
m2gh + 0 = 0 + m1gh + ½ Mv2 + ½m1v2 + ½m2v2
gh(m2 – m1) = ½( M + m1 + m2)v2
(10 m/s2)(1.0 m)(0.20 kg) = (0.625 kg)v2 v = 1.8 m/s
42.
Um – Wf = K'b + K'r-p + K'm
mmgh – mbgd = ½mbv2 + ½mpv2 + ½mmv2
(14)(10)(1) – (.25)(20)(10)(1) = (10 + 1 + 7)v2  v = 2.2 m/s
43.
L = rmv
L = (1 m)(1)(0.2 kg) (9 m/s) = 1.8 kg•m2/s
44. a.
L = rmv = r(2/5)m(2r/T) = 4mr2/5T
L = 4(6.0 x 1024 kg)(6.4 x 106 m)2/(5)(60 x 60 x 24 s)
L = 7.1 x 1033 kg•m2/s
b.
L = rmv = r(1)m(2r/T) = 2mr2/T
L = 2(6.0 x 1024 kg)(1.5 x 1011 m)2/(60 x 60 x 24 x 365 s)
L = 2.7 x 1040 kg•m2/s
45.
r1v1 = r2v2
v1/v2 = r2/r1 = (5.3 x 1012 m)/(8.9 x 1010 m) = 60
46.
rC CmCvC + rMMmMvM = (rCCmC + rMMmM)v'
(1)(42 kg)(3 m/s) + 0 = [(1)(42 kg) + (½)(180 kg)]v'
v' = 0.95 m/s
47. a.
Ltotal = rddmdvd + rr rmrvr
Ltotal = (R)(½)(M)(V) + (R)(1)(M)(0) = ½RMV
b.
Ltotal = Ltotal' = (rddmd + rrrmr)v'
½RMV = (½RM + RM)v'  v' = ⅓V
48. a.
mgh = ½mv2
v = (2gh)½ = [(2)(10 m/s2)(20 m)]½ = 20 m/s
b.
rmTvT + rmJvJ = (rmT + rmJ)v'
(100 kg)(20 m/s) + 0 = (145 kg)v'  v' = 13.8 m/s
c.
mgh = ½mv2
h = v'2/2g = (13.8 m/s)2/(2)(10 m/s2) = 9.5 m
d.
No
e.
Start from a higher ledge or with a running start.
f.
v' = (2gh)½ = [(2)(10 m/s2)(10 m)]½ = 14 m/s
rmTvT = r(mT + mJ)v'
vT = (145 kg)(14 m/s)/100 kg = 20.3 m/s
h = v2/2g = (20.3 m/s)2/(2)(10 m/s2) = 20.6 m
g.
½mv2 + mgh = ½mv'2
½v2 + (10 m/s2)(20 m) = ½(20.3)2  v = 3.5 m/s
49. a.
Ug = Krolling  mgh = ½(1 + ½)mv2
(10 m/s2)(1 m) = ¾v2  v = 3.65 m/s
b.
d = ½(v + vo)t
2 m = ½(3.65 m/s + 0)t t = 1.1 s
c.
Alteration
Final Velocity
Time
same
same
A 2.0-kg disk is used
A 1.0-kg ring ( = 1) is used
less
more
A 3-m ramp is used, but h = 1 m
same
more
50. a.
K' = Ug – Wf = mmgh – mbgd
K' = (5 kg)(10 m/s2)(1 m) – (0.3)(10 kg)(10 m/s2)(1 m) = 20 J
b.
K' = K'b + K'r-p + K'm = ½mbv2 + ½mpv2 + ½mmv2
20 J = (5 kg + 0.25 kg + 2.5 kg)v2 v = 1.6 m/s
51.
e.
vS + vS' = vB + vB'
6.3 m/s + vS' = 0 + vB'  vS' = vB' – 6.3
rmSvS + rmBvB = rmSvS' + rmBvB'
(1 kg)(6.3 m/s) = (1 kg)(vB' – 6.3) + (2 kg)vB' vB' = 4.2 m/s
vS' = vB' – 6.3 = 4.2 m/s – 6.3 m/s = -2.1 m/s
f.
hB = v'2/2g = (4.2 m/s)2/(2)(10 m/s2) = 0.88 m
hS = v'2/2g = (-2.1 m/s)2/(2)(10 m/s2) = 0.22 m
g.
mSghS + mBghB = (1 kg)(10 m/s2)(2 m) + 0 = 20 J
mBgh'B + mSgh'S = (2)(10)(.88) + (1)(10)(.22) = 19.8 J yes
55.
D
56.
A
104
1012
(8.9 x
m)(3.88 x
m/s) = (5.3 x
m)v2
v2 = 652 m/s
52. a.
L = rmv = r(2/5)m(2r/T) = 4mr2/5T
L = 4(7.35E22 kg)(1.74E6 m)2/(5)(2.42E6 s) = 2.3E29 kg•m2/s
b.
L = rmv = r(1)m(2r/T) = 2mr2/T
L = 2(7.35E22 kg)(3.84E8 m)2/(2.42E6 s) = 2.8E34 kg•m2/s
53.
rsmsvs + rMmMvM = (rsms + r mmm)v'
(1)(75)(5) + (½)(150)(2)= [(1)(75) + (½)(150)]v' v' = 3.5 m/s
54. a.
Ug = K  msgh = ½msvs2
vs = (2gh)½ = [2(10 m/s2)(2.0 m)]½ = 6.3 m/s
b.
rmSvS + rmBvB = r (mS + mB)v'
(1 kg)(6.3 m/s) + 0 = (1 kg +2 kg)v'  v' = 2.1 m/s
c.
Ug = K  (mS + mB)gh = ½(mS + mB)v'2
h = v'2/2g = (2.1 m/s)2/(2)(10 m/s2) = 0.22 m
d.
v' = mBvB/(mB + mS) = 12.6 kg•m/s/3 kg = 4.2 m/s
h = v'2/2g = (4.2 m/s)2/(2)(10 m/s2) = 0.88 m
You end where you started.
57.
D
Each A is ¼T  6 x ¼T = 3/2T.
58.
D
When x = 0, a = 0, but v = max; when x = A, v = 0, but a
= max; at intermediate points a  0 and v  0.
59.
B
E = ½kA2  mass will not change the total energy.
60.
D
E = ½kA2, whereas v = A(k/m)½ and a = A(k/m) change
the same, and T = 2(m/k)½ doesn't change at all.
61.
C
T = 2(m/k)½: when m doubles T increases by 2.
62.
C
T = 2(L/g)½: only changing L or g can change T.
63.
A
1010
Starting at 0: down A, up A, up A, down A.
T = 2(L/g)½: greater L = greater T.
64.
A
T = 2(L/g)½: to speed up the clock (reduce T), you
need to decrease L by moving the weight up.
65.
C
T = 2(L/g)½: "g" increases when accelerating up  T
decreases.
66.
A
T = 2(m/k)½: adding mass or changing the spring will
change the period.
67.
B
T = 2(L/g)½: amplitude has no effect on T.
68.
C
T = 2(L/g)½: Standing reduces the distance L (from
fulcrum to center of mass), which decreases T.
69.
B
T = 2(L/g)½: mass has no effect on T.
70. a.
Acceleration
Velocity
Maximum up
½T
Zero
¼ T, ¾ T
Maximum down
0 T, 1 T
¾T
0 T, ½ T, 1 T
¼T
b.
Practice Multiple Choice
amax
aA = A(k/m) = (0.1 m)(100 N/m)/(1 kg) = 10 m/s2
vmax
vo = A(k/m)½ = (0.1 m)[(100 N/m)/(1 kg)]½ = 1 m/s
T
T = 2(m/k)½ = 2(1/100)½ = 0.63 s
1.
C
2.
A
Kmax
Umax
Ko =
½mv2
UA =
½kA2
= ½(1 kg)(1
m/s)2
= (100 N/m)(0.1
= 0.5 J
m)2
3.
= 0.5 J
A
d.
¼T
0m
½T
-0.1 m
¾T
0m
1T
0.1 m
v
-1 m/s
0 m/s
1 m/s
0 m/s
5.
a
0 m/s2
10 m/s2
0 m/s2
-10 m/s2
A
T = 2(L/g)½ = 2(1.0 m/10 m/s2)½ = 2.0 s
m/s2
L=4m
4.0 s
g = 40
1.0 s
Maximum up
Zero
¾T
0 T, ½ T, 1 T
0 T, 1 T
Velocity
b.
Maximum down
¼T
¼ T, ¾ T
½T
aA = A(k/m) = (0.25 m)(100 N/m)/(1 kg) = 25 m/s2
vmax
vo = A(k/m)½ = (.25 m)[(100 N/m)/(1 kg)]½ = 2.5 m/s
T
T = 2(m/k)½ = 2(1/100)½ = 0.63 s
Kmax
Ko = ½mv2 = ½(1 kg)(2.5 m/s)2 = 3.125 J
Umax
UA = ½kA2 = ½(100 N/m)(0.25 m)2 = 3.125 J
¼T
+0.25 m
½T
0m
¾T
-0.25 m
1T
0m
v
0 m/s
-2.5 m/s
0 m/s
2.5 m/s
-25
0
25
m/s2
F
-25 N
0N
25 N
e.
maximum acceleration maximum velocity
73. a.
Ug = K  mgh = ½mv2
v = (2gh)½ = [2(10 m/s2)(0.20 m)]½ = 2 m/s
b.
T = 2(L/g)½ = 2(2.0 m/10 m/s2)½ = 2.8 s
The center of mass is the balance point, which is
closer to the more massive side.
7.
B
The released ball will move tangent to the circle at
that spot.
8.
A
The force of gravity is the only force acting on the
ball and Fg decreases as r increases (Fg = GMm/r2).
g = GM/r2, r is doubled  g is (½)2 = ¼ as much.
10.
Fc = mv2/r = (5.0 kg)(2.0 m/s)2/0.70 m = 29 N
11.
B
F = F
250 N + FL = 125 N + 500 N  FL = 375 N
12.
B
rFperson + rFboard = rFright chain
r(500 N) + (2 m)(125 N) = (4 m)(250 N) 
r = 1.5
13.
A
B
t
x
m/s2
At Q, acceleration is , at R acceleration is .
C and D would cancel torque, B would cancel up and
down, but only A will cancel both.
14.
d.
a
B
0 m/s2
0N
rFperson = rFplank (rplank is 2.5 m – 2 m = 0.5 m)
r(50 kg)(10 m/s2) = (0.5 m)(100 kg)(10 m/s2)  r = 1 m
15.
B
m/s2
Ball simultaneously is directed to the right (due to
the spring) and toward the top (due to the rotation).

6.
B
72. a.
amax
D
D
Acceleration
Fg = GMm/r2, where r is doubled  Fg should be
(½)2 = ¼ as much  ¼(800 N) = 200 N.
9.
c.
m = 4 kg
2.0 s
Traveling east and turning south  clockwise.
ac = v2/r  v = (acr)½ = [(3 m/s2)(300 m)]½ = 30 m/s
4.
t
x
F
0N
10 N
0N
- 10 N
e.
Max acceleration
Max velocity
Period
doubles
doubles
remains the same
71. a.
Ug = K  mgh = ½mv2
v = (2gh)½ = [2(10 m/s2)(0.015 m)]½ = 0.55 m/s
b.
In uniform circular motion velocity is tangent to the
circle and acceleration is toward the center.
g = GM/r2, where r is doubled
 g is (½)2 = ¼ as much. ¼(10 m/s2) = 2.5 m/s2
16.
C
17.
C
Fg = GMm/r2, where r is doubled
 Fg is (½)2 = ¼ as much.
CM =  (rimi)/(mi)—assume the rod is 4 m
2 = [(0)(3.5) + (1)(X) + (4)(5)]/(8.5 + X)  X = 3
18.
period
D
T = Fc = mv2/r = mv'2/r': r increases to 4r without
changing T  v'2 also increase by 4. v'2 = 4  v' = 2v.
19.
A
20.
C
At the bottom of the swing, Fnet = Fc = Ft – Fg
Fc = 6 N – (0.4 kg)(10 m/s2) = 2 N
net = cc – c
net = (3R)F + (3R)F + (2R)F – (3R)(2F) = 8RF – 6RF = 2RF
21.
B
g = GM/r2  gE = GME/rE2 and GM = GMM/rM2
gM/gE = (MM/ME)(rE/rM)2 = (1/10)(1/½)2 = 4/10  gM = 2g/5
22.
C
44.
Fc = Fg
mv2/r = GMm/r2  r = GM/v2
23.
C
I1m = l2M1 and l1M2 = l2m  l1/l2 = M1/m = m/M2 
M1M2 = m2  m = (M1M2)½
25.
D
Fg = Fs  mg = kd  k = mg/d
T  (m/k)½ = (md/mg)½ = (d/g)½
45.
Rm + (rcos60)M = R(2M)
m + ½M = 2M  m = 3/2M
24.
D
A
D
The 4-kg block stretches the spring 16 cm (Fs  x),
the spring falls another 16 cm before turning around.
46.
D
v = A(k/m)½
v2 = A2k/M  k = Mv2/A2 = M(v/A)2
47.
Energy & angular momentum are conserved,
r  v  (r1v1 = r2v2), Ug increases (Ug = -GMm/r).
D
Tp = 2(L/g)½ = Ts = 2(m/k)½
L/g = m1/k  k = m1g/L
26.
A
1.
27.
D
Practice Free Response
r1v1 = r2v2  v2 = (r1/r2)v1
a. (1)
I is true (Fg = Fc =
II is true (F r  no work
done), and III is true (angular momentum is constant).
Mv2/R),
28.
B
A has no force, C and D have a constant force, only
SHM has a variable force ( Fs = kx).
29.
A
Ug = K (no rolling because of zero friction)
mgh = ½mv2 v = 2gh = (2gh)½
P
(2)
Ug = K  mgh = ½mv2
v = (2gh)½ = [2(10 m/s2)(90 m)]½ = 42 m/s
b.
30.
D
mgh = K + Kr = ½mv2 + ½mv2 = ½(1 +  )mv2
v2 = 2gh/(1 + 2/5)  v = (10gh/7)½
31.
B
v = (2gh)½ = [2(10 m/s2)(40 m)]½ = 28 m/s
L = rmv = (a)(1)(m)(v) = amv
c.
(1)
32.
C
Total mechanical energy is constant when a
conservative force is involved.
Fg
33.
D
The kinetic energy is greatest when the velocity is
greatest (K = ½mv2), which occurs at the midpoint.
34.
B
Potential energy is greatest when x is greatest
(Ug = ½kx2), which occurs at xmin and xmax,
2.
Acceleration is greatest when force is greatest
(Fs = kx), which occurs at xmin and xmax.
b.
Velocity is greatest at the midpoint and is zero at xmin
and xmax (same as K).
c.
Fg = mmg = (10 kg)(10 m/s2) = 100 N
36.
D
Fnet = 100 N – 60 N = 40 N
37.
D
38.
D
T = 2(m/k)½  T depends on both m and k, but not
on amplitude, A.
d.
Fnet = (mb + mp + mm)a
40 N = (20 kg + 5 kg + 10 kg)a  a = 1.1 m/s2
e.
v2 = vo2 + 2ad
v2 = 02 + 2(1.1 m/s2)(2 m)  v = 2.1 m/s
f.
T = 2(m/k)½ T m.
The mass is doubled  T = 22 = 2.8 s.
39.
D
The period of a pendulum, T = 2(L/g)½  T  (1/g)½
and it is shorter, but the spring period is the same.
Ug = mmgh = (10 kg)(10 m/s2)(2 m) = 200 J
40.
C
Potential energy is greatest when x is greatest
(Ug = ½kx2), which occurs at 1 s and 3 s.
g.
Each spring supports half of the weight (12 N).
Fs = kx  6 N = k(0.15 m)  k = 40 N/m.
h.
Um – Wf = K'b + K'r-p + K'm = ½mbv2 + ½mpv2 + ½mmv2
200 J – 120 J = ½(20 kg + 5 kg + 10 kg)v2  v = 2.1 m/s
a.
Ug = mg(H + Lsin)
Ug = (0.5 kg)(10 m/s2)[1 m + (2 m)(sin30)] = 10 J
Wf = mbgd = (.30)(20 kg)(10 m/s2)(2 m) = 120 J
41.
A
42.
D
The force on ball is constant, except when it is in
contact with the floor,  not SHM, where force  x.
43.
B
(2)
Fc = mv2/r = (700 kg)(28 m/s)2/(20 m) = 27,000 N
Fg = mg = (700 kg)(10 m/s2) = 7,000 N
Fc = Fg + Fn  Fn = Fc – Fg = 27,000 – 7,000 = 20,000 N
a.
F = mbg = (0.30)(20 kg)(10 m/s2) = 60 N
35.
B
Fn
Pendulum: T = 2(L/g), T = T
Spring: T = 2(m/k), where T m  T = 2T
3.
b. (1)
Ug = Krolling mgLsin = ½(1 +  )mv2
(10 m/s2)(2 m)sin30 = ½(1 + 1)v2  v = 3.2 m/s
(2)
Kfloor = Ktable + Utable  ½mvf2 = ½mvt2 + mgh
vf2 = (3.2 m/s)2 + 2(10 m/s2)(1 m) vf = 5.5 m/s
(3)
5.
T = 2(L/g)½ = 2(1.0 m/10 m/s2)½ = 2.0 s  1.0 s
c.
K = 0.4 J – U = 0.4 J – 0.2 J = 0.2 J
K = ½mv2 = ½(0.5 kg)(5.5 m/s)2 = 7.5 J
d.
(4)
K = ½mv2  0.4 J = ½(3.0 kg)v2  v = 0.5 m/s
(10 J – 7.5 J)/10 J x 100 = 25 %
c. (1)
The hoop (largest  = largest % rotational energy)
(2)
The disk (smallest  = most translational energy)
(3)
Both would have the same amount of kinetic energy.
4.
t
x (m)
v (m/s)
a (m/s2)
F (N)
U (J)
K (J)
a.
10 cm
b.
0s
0
–
0
0
0
+
1s
–
0
+
+
+
0
2s
0
+
0
0
0
+
3s
+
0
–
–
+
0
6.
a.
Ug = 2mgh
Ug = 2(1 kg)(10 m/s2)(0.50 m) = 10 J
b.
Ug = K = ½mv2 + ½(1 +  )mv2 = 5/4 mv2
10 J = 5/4(2 kg)v2 v = 2.8 m/s
c.
mcvc + mbvb = (mc + mb)v'
2mvc + 0 = 5mv'  v' = 2/5vc = 1.1 m/s
d.
Kt' = ½mv2
Kt'= ½(5 kg)(1.1 m/s)2 = 3.2 J
e.
K = Us = ½kx2
3.2 J = ½(250 N/m)x2  x = 0.16 m