Sample Practice Questions
1. A 1,000 kg rocket ship, traveling at 100 m/s, is acted on by an average force of 20 kN
applied in the direction of its motion for 8 s. What is the change in velocity of the
rocket?
A. 100 m/s
B. 160 m/s
C. 1,600 m/s
D. 2,000 m/s
2. An elevator is designed to carry a maximum of weight of 9,800 N (including its own
weight) and to move upward at a speed of 5 m/s after an initial period of
acceleration. What is the relationship between the maximum tension in the elevator
cable and the maximum weight of 9,800 N when the elevator is accelerating upward?
A. The tension is greater than 9,800 N.
B. The tension is less than 9,800 N.
C. The tension equals 9,800 N.
D. It cannot be determined from the information given.
3. A 10 kg wagon rests on a frictionless inclined plane. The plane makes an angle of 30°
with the horizontal. What is the force, F, required to keep the wagon from sliding
down the plane?
A. 10 N
B. 30 N
C. 49 N
D. 98 N
4. A 20 kg wagon is released from rest from the top of a 15 m long plane, which is
angled at 30° with the horizontal. Assuming there is friction between the ramp and
the wagon, how is this frictional force affected if the angle of the incline is increased?
A. The frictional force increases.
B. The frictional force decreases.
C. The frictional force remains the same.
D. It cannot be determined from the information given.
5. An astronaut weighs 700 N on Earth. What is the best approximation of her new
weight on a planet with a radius that is two times that of Earth, and a mass three
times that of Earth?
A. 200 N
B. 500 N
C. 700 N
D. 900 N
6. A 30 kg girl sits on a seesaw at a distance of 2 m from the fulcrum. Where must her
father sit to balance the seesaw if he has a mass of 90 kg?
A. 67 cm from the girl
B. 67 cm from the fulcrum
C. 133 cm from the girl
D. 267 cm from the fulcrum
7. An object is moving uniformly in a circle whose diameter is 200 m. If the centripetal
acceleration of the object is 4 m/s2, then what is the time for one revolution of the
circle?
A. 10π s
B. 20π s
C. 100 s
D. 200 s
8. A distant solar system is made up one small planet of mass 1.48 × 1024 kg revolving in
a circular orbit about a large, stationary star of mass 7.3 × 1030 kg. The distance
between their centers is 5 × 1011 m. What happens to the speed of the planet if the
distance between the star and the planet increases (assume the new orbit is also
circular)?
A. The speed of the planet increases
B. The speed of the planet decreases.
C. The speed of the planet remains the same.
D. It cannot be determined from the information given.
9. An elevator is accelerating down at 4 m/s2. How much does a 10 kg fish weigh if
measured inside the elevator?
A. 140 N
B. 100 N
C. 60 N
D. 50 N
10. A 1 kg ball with a radius of 20 cm rolls down a 5 m high inclined plane. Its speed at
the bottom is 8 m/s. How many revolutions per second is the ball making when at
the bottom of the plane?
A. 6 revolutions/second
B. 12 revolutions/second
C. 20 revolutions/second
D. 23 revolutions/second
Explanations to Practice Questions
1. B
The average force on the rocket equals its mass times the average acceleration; the
average acceleration equals the change in velocity divided by the time over which the
change occurs. So the change in velocity equals the average force times the time
divided by the mass:
F = ma, so a = F/m
a = Δv/Δt
Δv = aΔt = FΔt/m
Δv = (20,000 N × 8 s)/1,000 kg
Δv = 160 m/s
(B) is therefore the correct answer.
2. A
The forces on the elevator are the tension upward and the weight downward, so the
net force on the elevator is the difference between the two. When the elevator is
accelerating upward, the tension in the cable must be greater than the maximum
weight so that the difference produces a nonzero acceleration. (A) is therefore the
correct answer. For the sake of completion, when the elevator is moving at a
constant speed, its acceleration is zero, so the net force on the elevator is also zero,
which means that the tension and the weight are equal. Similarly, when the elevator is
accelerating downward, the tension is less than 9, 800 N so that there is a nonzero
acceleration.
3. C
The weight of the wagon (W= mg) acts in a direction straight down. This force can
be separated into X and Y components, with the x-axis parallel to and the y-axis
perpendicular to the plane of the incline.
Wx = W sin 30°
Wy = W cos 30°
To keep the wagon from moving (i.e., to keep it in equilibrium), the sum of the
forces must equal zero. In terms of the components,
ΣFx =0 and ΣFy= 0
From a free-body diagram, it can be seen that the Y component of the weight is
counteracted by an equal and opposite force from the surface of the plane (the
normal force, N). Similarly, the X component of the weight, which tends to cause the
wagon to roll down the plane, must be counteracted by an equal and opposite force
parallel to the plane. This is the unknown force, F. (Remember, the plane is
frictionless.) The magnitude of the required force can be determined by recognizing
that the sum of the forces in the x-direction must equal zero:
(C) best matches our result and is thus the correct answer.
“down the plane” to be the positive x-direction, as long as we use the correct signs
when we calculate the components. The choice of positive direction is arbitrary.
4. B
The force of friction on an object sliding down an incline equals the coefficient of
friction times the normal force. The normal force, which is given by mg cos θ
decreases as the angle of the incline, θ increases. Therefore, the frictional force
decreases as the angle of the incline increases. (B) best reflects this and is thus the
correct answer.
5. B
The weight of an object on a planet can be found by using Newton’s law of
gravitation with one minor simplification. Because the planet is much larger than the
person, r, which is the distance between the centers of the two objects (the planet
and person), can be taken simply as the radius of the planet; the additional few
meters to the center of the astronaut can be ignored. The astronaut’s weight on earth
equals
F= GmME/rE
2 = 700 N
On a planet that is three times as massive as earth (Mp = 3ME), and has twice the
radius (rP = 2rE), the force would be equal to
F = GmMP/r2
p
F = Gm (3ME)/(2rE)2
F = 3GmME/(4rE
2)
F = ¾ (GmME/rE
2)
F = ¾ (the astronaut’s weight on earth)
F = ¾ × 700N
F 500 N
(B) is therefore the correct answer.
6. B
For the seesaw to be balanced, the torque due to the girl must be exactly
counteracted by the torque due to her father. More generally stated, the sum of the
torques about any point must be equal to zero. Taking the torques about the fulcrum
for the girl and the father, obtain the following (Note, the f subscript represents the
father while the g subscript represents the girl):
torques about any point, not just the fulcrum. Doing so, however, is usually more
complicated because the force of the fulcrum on the seesaw first has to be
determined. Because both the weight of the girl and the weight of the father act in a
downward direction, the fulcrum must exert a force of (30 × g) + (90 × g) = 120 g N
in an upward direction on the seesaw. Calculating the torques about the girl’s
location, we obtain the following:
This question is testing your knowledge of circular motion. To calculate the time for
one revolution of the circle, start with the centripetal acceleration: a = v2/r. Given r =
100 m and a = 4 m/s2, we have v = 20 m/s. Speed, v, and time of revolution (period),
T, are related by v = 2πr/T (i.e., in a time equal to one revolution, the distance
covered is the circumference). Therefore,
T= 2π (100 m)/(20 m/s) = 10πs
(A) matches your result and is therefore the correct answer.
8. B
The force on the planet due to the star is equal to
F = Gm starmplanet/r2
Because this force provides the centripetal force, it’s also true that F is equal to
F= mplanetv2/r
Equating the two expressions for F gives the following:
decreases. This means that (B) is the correct answer.
9. C
You may imagine that the fish is attached to a string and the tension in the string is
given by ma = mg - T (because a and g point in the same direction). Then T = m ×
(g - a) =10 × (10 - 4) = 60 N. A quicker way to solve this problem is to notice that
when the elevator accelerates down with magnitude g, the fish is in free fall and
weighs 0 N. Alternatively, when the elevator goes up, one usually feels heavier. (C) is
therefore the correct answer.
10. A
In 1 second, the ball passes 8 m. One revolution is 2πr = 1.3 m, so 8/1.3 is 6
revolutions/s. Therefore, (A) is the correct answer.
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