b.
Practice Problems
mtotal = 1 kg + 1 kg + 0.05 kg + 2.05 kg
1.
B
c.
Fnet = ma
0.500 N = (2.05 kg)a  a = 0.244 m/s2
d.
d = ½at
1.00 m = ½a(2.90 s)2  a = 0.238 m/s2
e.
Fg and Fn cancel each other out.
2.
C
Fnet = ma = m(0) = 0
3.
B
Friction slows the book  force is backward.
%  = 100(0.244 – 0.238)/0.244 = 2 %
4.
B
Friction slows the stone  force is backward.
A
5.
B
6.
B
Fnet = 0  no acceleration and the speed stays the
same.
Ft = mv, but mv is constant  Ft is constant.
Ft = ½Ft'  t' = 2t
A
Ft = mv, double t doubles v, but v2 = 2ad  double
t quadruples d.
F = m1a1 = m2a2 and a2 = 2a1 m2 = ½m1.
mtot = m1 + ½m1 = 3/2m1  a = 2/3a1
B
Both fell at the same rate  same acceleration.
Both have equal accelerations, but the hammer has
greater mass (Fg = ma).
23.
Friction can only resist motion, not cause it, since the
horizontal force < force of friction, then no motion.
24. a.
Fp- = Fpsin30 = (100 N)sin30 = 50 N
12.
C
F – Ff = ma, since Ffa > Ffb, then less force is available
for acceleration  aa < ab.
C
11.
B
Since Fa = Ffa and Fb = Ffb, but Ffa > Ffb Fa > Fb.
22.
10.
C
Ffa = Na and Ffb = Nb, but Na > Nb  Ffa > Ffb.
21.
A
9.
C
Na = mg + Fsin and Nb = mg – Fsin  Na > Nb.
20.
8.
A
Newton's third law reaction force generates the
forward force.
19.
A
Ft = mv, double v doubles t.
NA = Fg, but NB = Fgcos, which is less than Fg.
18.
B
7.
B
17.
F = ma: with the same F and m, a will be the same.
b.
Fp- = Fpcos30 = (100 N)cos30 = 87 N
13.
car is accelerating northward
northward
bowling ball rolling straight at constant speed
zero
thrown rock reaches its highest point
downward
rock resting on the ground
zero
14.
30 N
30 N
Fnet = -30 N + 30 N = 0 N
Fnet = ma
0 N = (10 kg)a  a = 0 m/s2
30 N
50 N
Fnet = -30 N + 50 N = 20 N
Fnet = ma
20 N = (10 kg)a  a = 2 m/s2
30 N
20 N
Fnet = -30 N + 20 N = -10 N
Fnet = ma
-10 N = (10 kg)a  a = -1m/s2
15. a.
Third Law
b.
First Law
16. a.
Fnet = mg = (0.0500 kg)(10 m/s2) = 0.500 N
c.
F|| = ma
87 N = (10 kg)a  a = 8.7 m/s2
25. a.
Fg = mgearth = (75 kg)(9.8 m/s2) = 735 N
b.
Fg = mgmoon = (75 kg)(1.6 m/s2) = 120 N
26. a.
Fn = Fg = mg = (10 kg)(10 m/s2) = 100 N
b.
Fn = Fg + Fp = 100 N + 25 N = 125 N
c.
Fn = Fg – Fp = 100 N – 25 N = 75 N
d.
Fn = Fgcos= (100 N)cos30 = 87 N
27. a.
F = Ff = sFn = (0.3)(10 kg)(10 m/s2) = 30 N
b.
Ff = 10 N
c.
Ff = kFn = (0.15)(100 N) = 15 N
28.
D
e.
Ff = Fn
0.3 N = (5 N)   = 0.06
41. a.
Force is proportional to mass when acceleration is the
same, since m = 1/3 total mass then force = 1/3F.
Fn
Ff
29.
A
T1 pulls all the mass, T2 pulls 5/6 of the mass and T3
pulls ½ of the mass, T1 > T2 > T3.
30.
C
Newton's third law of equal and opposite forces.
31.
C
Whether tied to a tree or pulled with equal force, the
tension is the same, 100 N.
32.
D
The combined forces on one end will generate a
reaction force equal to 200 N
33.
B
F = ma, where m = 50 kg + 10 kg for case 1 and 50 kg
for case 2  case 2 where the mass is less.
34.
C
35.
C
F|| = Fgsin – Fgcos: as  increases the first term
increases and the second term decreases.
F|| = mgsin – mgcos= ma: m cancels out of the
equation  m doesn't affect the situation.
36.
B
Since acceleration is zero, Fnet = Fn – Fg = 0
 Fn = F g .
37.
A
Since acceleration is upward, Fnet = Fn – Fg > 0
Fn > Fg.
Fg
b.
Fg-|| = Fgsin
Fnet = (80 kg)(10 m/s2)sin30 = 400 N
c.
Fg- = Fgcos
Fg- = (80 kg)(10 m/s2)cos30 = 690 N
d.
Fn = Fg- = 690 N
e.
Ff = Fn
Ff = 0.10(690 N) = 69 N
f.
Fnet = Fg-|| – Ff
Fnet = 400 N – 69 N = 330 N
g.
Fnet = ma
330 N = (80 kg)a  a = 4.1 m/s2
42. a.
38.
Since acceleration is equal to -g, Fnet = Fn – Fg = -mg
Fn = Fg – mg = 0.
39. a.
Fp = 70 N
Ft-1
D
Ft-2
Fg = 50 N
b.
Fnet = Fp – Fg
Fnet = 70 N – (5 kg)(10 m/s2) = 20 N
c.
Fnet = ma
20 N = (5 kg)a  a = 4 m/s2
d.
vt = vo + at
v = 0 + (4 m/s2)(2 s) = 8 m/s upward
40. a.
Fn
Ff
Fg
b.
vt2 = vo2 + 2ad
(2 m/s)2 = (10 m/s)2 + 2a(80 m)  a = -0.6 m/s2
c.
Ff = ma
Ff = (0.5 kg)(-0.6 m/s2) = -0.3 N
d.
Fn = Fg = mg
Fn = (0.5 kg)(10 m/s2) = 5 N
Fg-1 = 11500 N
Fg-2 = 100,000 N
b.
Fnet = Fg-1 – Ft-1 + Ft-2 – Fg-2 = 11500 N – 10000 N = 1500 N
c.
Fnet = ma
1500 N = (m1 + m2)a = (2150 kg)a  a = 0.70 m/s2
d.
Fnet = Fg-1 – Ft-1 = m1a
11500 N – Ft-1 = (1150 kg)(0.70 m/s2)  Ft-1 = 10700 N
e.
Fnet = Ft-2 – Fg-2 = m2a
Ft-2 – 10000 N = (1000 kg)(0.70 m/s2)  Ft-2 = 10700 N
f.
d = vot + ½at2
3 m = 0 + ½(0.70 m/s2)t2  t = 2.9 s
g.
v2 = vo2 + 2ad = 0 + 2(0.70 m/s2)(3 m) = 4.2 m2/s2
v = 2.0 m/s
h.
Fnet = Fg – Fn = ma
500 N – Fn = (50 kg)(0.70)  Fn = 465 N
43. a.
Fp = 325 N
46. a.
Ff
Fn
 Fg
Fg
Fg-||

b.
Fg-|| = Fgsin
Fg-|| = (10 kg)(10 m/s2)sin37 = 60 N
c.
Fg- = Fgcos
Fg- = (10 kg)(10 m/s2)cos37 = 80 N
d.
Fg = 250 N
b.
Fnet = Fp – Fg = 325 N – 250 = 75 N
c.
Fnet = ma
75 N = (25 kg)a  a = 3 m/s2
d.
Fn = Fg- = 80 N
vt = vo + at = 0 + (3 m/s2)(2 s) = 6 m/s
44. a.
Fn
Ff
Ft-A
Ft-B
Fg-A
Fg-B
b.
Ff = Fn = mAg
Ff = (0.30)(10 kg)(10 m/s2) = 30 N
c.
Fnet = Fg-B – Ff
Fnet = (5 kg)(10 m/s2) – 30 N = 20 N
d.
Fnet = (mA + mB)a
20 N = (10 kg + 5 kg)a  a = 1.3 m/s2
e.
Fnet = Fg-B – Ft = mBa
(5 kg)(10 m/s2) – Ft = (5 kg)(1.3 m/s2)  Ft = 44 N
f.
vt2 = vo2 + 2ad
vt2 = 0 + 2(1.3 m/s2)(0.5 m)  vt = 1.1 m/s
45. a.
Fn
e.
Ff = Fn
Ff = 0.25(80 N) = 20 N
f.
Fnet = Fg-|| – Ff
Fnet = 60 N – 20 N = 40 N
g.
Fnet = ma
40 N = (10 kg)a  a = 4 m/s2
47. a.
Ft-1.5
Fg-1
Fg-1.5
b.
Fnet = Fg-1.5 – Ft-1.5 + Ft-1 – Fg-1
Fnet = (m1.5 – m1)g = (1.5 kg – 1 kg)(10 m/s2) = 5 N
c.
Fnet = ma
5 N = (2.5 kg)a  a = 2 m/s2
d.
Fnet = Fg-1.5 – Ft-1.5 = m1.5a
15 N – FT-1.5 = (1.5 kg)(2 m/s2)  Ft-1.5 = 12 N
e.
Fnet = Ft-1 – Fg-1 = m1a
Ft-1 – 10 N = (1 kg)(2 m/s2)  Ft-1 = 12 N
f.
d = vot + ½at2
2.25 m = 0 + ½(2 m/s2)t2  t = 1.5 s
Ff
Fg
b.
vt2 = vo2 + 2ad
(0 m/s)2 = (15 m/s)2 + 2a(90 m)  a = -1.25 m/s2
c.
Ff = ma
Ff = (0.50 kg)(-1.25 m/s2) = -0.625 N
d.
Fn = Fg = mg
Fn = (0.50 kg)(10 m/s2) = 5 N
e.
Ff = Fn
0.625 N = (5 N)   = 0.125
Ft-1
48.
a.
W + Ff = 75 N
W = 75 – Fg = 75 – (0.40)(100) = 35 N
b.
W = 75 + Ff
W = 75 + 40 = 115 N
49. a.
TLcos135 + TRcos30 = 0
TL(-0.707) + TR(0.866) = 0  TR = TL(0.816)
b.
TLsin135 + TRsin30 + mgsin-90 = 0
TL(0.707) + TL(0.816)(0.500) + (50 kg)(10 m/s2)(-1) = 0
TL(1.12) = 500 N  TL = 448 N
c.
TR = TL(0.816) = (269 N)(0.816) = 366 N
50. a.
11.
B
12.
A
TL
D
TR
b.
C
c.
51. a.
b.
Practice Multiple Choice
D
The vector sum forms a 6-8-10 (3-4-5) triangle, which
is a right triangle, where the 3-4 sides make 90o.
2.
D
Fg = mg
1560 N = (60 kg)g g = 26 m/s2
3.
C
Newton's third law states that for every force there is
an equal and opposite reaction force.
4.
A
Without acceleration then F = Ff.
Ff = Fn = mg Ff = (2/3)(60 N) = 40 N
5.
B
6.
A
Newton's third law states that for every force there is
an equal, but opposite reaction force.
Fs = kx  mg = kx
(10 kg)(10 m/s2) = (100 N/m)x  x = 1 m
7.
A
The coefficient k is less when the object is sliding
(k < s)  it takes less force to keep it sliding.
8.
C
9.
B
No acceleration then F = Ff = Fn 
40 N = (0.05)Fn  Fn = 800 N
Slowing down the force of friction, Ff, must be
greater than the push/pull force, F
10.
D
Fnet = ma (Fnet = extra weight and m = total mass)
(0.30 kg)(10 m/s2) = (.30 + .60 + .60 kg)a  a = 2 m/s2
Fnet = ma (Fnet is Fg on m, and m = total mass = 4 m)
mg = (4m)a  a = ¼ g
19.
Fnet = Ff = Fn
(50 N)cos37 = [(10 kg)(10 m/s2) – (50 N)sin37]  = 4/7
20.
Fnet = F1 – F2cos60 = ma (cos60 = ½)
50 N – 40 N(½) = (10 kg)a  a = 3 m/s2 (to the right)
21.
A
1.
Fnet = Ff + Fg-||
Fnet = 4/5mg) + 3/5mg = (0.3)(40 N) + 30 N = 42 N
18.
C
cos53 = TR/Fg = TR/500 N  TR = 300 N
Fnet = Fp – Fg = ma
50 N – (3 kg)(10 m/s2) = (3 kg)a  a = 20/3 = 6.7 m/s2
17.
D
sin53 = TL/Fg = TL/500 N  TL = 400 N
Fnet = ma = 0
Fp – Ff – Fgsin = 0  Fp = Ff + Fgsin
16.
B
tan60 = Fg/TR = 500 N/TR  TR = 289 N
Ff = Fn
15.
C
sin60 = Fg/TL = 500 N/TL  TL = 577 N
Fn equals the component of gravity that is
perpendicular to the surface: Fgcos.
14.
B
60o
Force is proportional to the mass being pushed 
F2/F = 2/5  F2 = 2/5F = 2/5(20 N) = 8 N (left).
13.
B
Fg
Fnet = F – Ff = ma 
50 N – Ff = (4.0 kg)(10 m/s2)  Ff = 10 N
22.
D
23.
D
Ff = Fn
Ff = (mg + Tsin)
Fnet = Fcos – f = ma
a = (Fcos – f)/m
f = Fn = (mg – Fsin)
 = f/(mg – Fsin)
24.
C
Fnet = Ff – Fg = 0
0 = Fn – mg = F – mg  F = mg/
25.
D
Fnet = ma
(M – m)g = (M + m)a  a = (M – m)g/(M + m)
26.
C
When you place the individual force vectors tail to
tip, only c will equal the net force vector.
27.
D
TL, Fg and TR make a 30-60-90 triangle, where
sin30 = Fg/TR  TR = (100 N)/sin30 = 100 N/½ = 200 N.
28.
C
tan30 = Fg/TL
 TL = (100 N)/tan30 = 100 N/(1/3) = 170 N
Practice Free Response
1.
5.
a.
a.
Fn
Ff
2.
Fg
b.
Fnet = Fgsin – Ff = 0
Ff = (25 kg)(10 m/s2)sin15o = 65 N
c.
Ff = Fn = (Fgcos)
65 N = (25 kg)(10 m/s2)cos15o  = 0.27
a.
d = vot + ½at2
55 m = (25 m/s)(3.0 s) + ½a(3.0 s)2 a = -4.4 m/s2
b.
Fn
Ff
3.
4.
Fg
c.
F = Ff = Fn
ma = mg   = a/g = 4.4 m/s2/10 m/s2 = 0.44
d.
a = (v – vo)/t = 25 m/s/10 s = 2.5 m/s2
F = Fs = kx  ma = kx
x = ma/k = (900 kg)(2.5 m/s2)/(9200 N/m) = 0.24 m
e.
The spring shortens. Without acceleration, the spring
force is not needed to keep the crate on the truck bed
so the spring will go to its unstretched length.
a.
Fg = Ff  (mbucket + msand)g = smblockg
1.35 kg + msand = (0.450)(28.0 kg)  msand = 11.25 kg
b.
Fnet = Fgbucket + Fgsand – Ffblock = mtotala
(mbucket + msand – kmblock)g = (mbucket + msand + mblock)a
a = [12.6 kg – (0.320)(28.0 kg)]10 m/s2]/40.6 kg = 0.90 m/s2
a.
Fp = mg + ma = m(g + a)
Fp = (7650 kg + 1250 kg)(10.80 m/s2) = 96,100 N
b.
Fnet = ma  Ft – mg = ma
Ft = m(a + g) = (1250 kg)(0.80 m/s2 + 10 m/s2) = 13,500 N
c.
Fp – Fg –Ft = ma
96,100 N – 76500 N – Ft = (7650 kg)(0.80 m/s2)
Ft = 13,500 N
TB
TA
Fg-B
Fg-A
b.
Fn = Fg-A – Fg-B = mAg – mBg = (mA – mB)g
Fn = (70 – 60)(10) = 100 N
c.
Fnet = TB – Fg-B = ma
TB = mBg + mBa = mB(g + a) = (60)(10 + 0.25) = 615 N
d.
No, the upward force TA = 615 N < the downward force
Fg-A = 700 N  the net force on student A is downward.
e.
Fnet = TB – FgB = mBa (TB = TA = mAg)
mAg – mBg = mBa
a = (mA – mB)g/mB = (70 – 60)10/60 = 1.7 m/s2
f.
Anchor student A with additional weight.