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Solution’s to Newton’s Laws Exam Review
1. First produce an FBD.
FN
Ff
Fapp
Fg
a. The FNet = 0 so the Fapp = Ff , therefore the force of friction is 30 N. Also the
normal force is equal to the weight. (Fg = mg)
Ff = k FN
30 N = k (12kg x 9.8 m/s2)
k = 0.255
b. The normal force in this case is equivalent to his weight which is 560 N because
he is in static equilibrium (at rest). Therefore, the FN = 560 N.
c. At rest (static equilibrium) or moving at constant velocity (dynamic equilibrium).
2. a. Applied force, tension, weight (or force of gravity), Normal force, Thrust, Friction,
Drag force
b. Find FNET : FNET = ma , FNET = (800 N / 9.8 m/s2) x 5 m/s2 = 408 N
FN
Fg
FNET = FN – 800 N = 408 N, so the force exerted by the elevator is FN = 1208 N
c. The reaction force from the ground pushes the runner forward with a force
equivalent to 550 N. The net force on the runner will be 80 kg x 1.5 m/s2 = 120 N
The difference between the 550 N force and the drag force must be equal to 120 N
forward and therefore the drag force is equal to 430 N.
3. a. Acceleration varies directly. (FNET increases and acceleration increases.)
b. Acceleration varies inversely. (Mass increases and acceleration decreases.)
c. Use kinematics equation:
a
v f  vi
t

2.5m / s  0
m
5 2
0.5s
s
Now use a = FNET/m
FNET = 1.5 kg (5 m/s2) = 7.5 N
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Solution’s to Newton’s Laws Exam Review
4. a. This is wood on wood. Use your chart in order to find the static coefficient of
friction. s = 0.42
b. Now we set Ffs = s FN Because the floor is horizontal, the normal force is equal
to the weight. So,
Ffs = 0.42 (2kg x 9.8 m/s2) = 8.23 N
c. TRICKY… Because the force has not met or exceeded the maximum force of
friction, the force of friction must be equal to the applied force if the object will
remain at rest. There must be a net force of zero. Therefore, Ff = 8 N.
d. Now the applied force has exceeded the max static force of friction and the shoe
begins to slide. Kinetic friction is now present. The object will accelerate.
Challenge yourself! Figure out the acceleration of the shoe. (Answer: 4.6 m/s2)
5. a. The normal force and the force of gravity.
b. Parallel (Fg sin  ), Perpendicular ( Fg cos  )
c. FNET = 0
(Wparallel = Tension )
mgsin  = 32 N
m (9.8 m/s2) sin 530 = 32 N
m = 4.09 kg
d. The sofa moves up at constant velocity, FNET = 0
FNET = Fapp – Ff – mgsin ,
Fapp = Ff + mgsin 
= 0.9 (60 kg x 9.8 m/s2 cos 250) + 60 kg x 9.8 m/s2 sin 250)
= 479.6 + 248.5 = 728 N
6. Newton’s 3rd Law is for every force there is an equal in magnitude and opposite in
direction reaction force. The two forces act on different objects.
If the girl pushes on the boat with 100 N, the boat pushes on the girl with 100 N.
Since the girl has less mass, the girl will accelerate at a greater rate.
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