Dynamics Practice Questions
AP (Advanced Placement) · AP Physics 1 · 149 free MCQs with instant results and detailed explanations.
149
Total
53
Easy
66
Medium
30
Hard
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Sample Questions from Dynamics
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Q1
Easy
A 2 kg object is subjected to a net force of 10 N. What is the acceleration of the object?
A.
2 m/sยฒ
B.
4 m/sยฒ
C.
5 m/sยฒ
D.
10 m/sยฒ
Show Answer & Explanation
Correct Answer: C
Using Newton's second law, F = ma, we can rearrange it to find acceleration a = F/m. Here, a = 10 N / 2 kg = 5 m/sยฒ. Therefore, the correct answer is C.
Q2
Easy
A car accelerates from rest at a constant rate of 2 m/sยฒ for 5 seconds. What is the final velocity of the car?
A.
10 m/s
B.
15 m/s
C.
5 m/s
D.
20 m/s
Show Answer & Explanation
Correct Answer: A
Using the formula v = u + at, where u is the initial velocity (0 m/s), a is acceleration (2 m/sยฒ), and t is time (5 s), we find v = 0 + (2)(5) = 10 m/s.
Q3
Easy
Which of the following best describes the relationship between mass and acceleration in Newton's second law of motion?
A.
Mass is inversely proportional to acceleration.
B.
Mass is directly proportional to acceleration.
C.
Acceleration is constant regardless of mass.
D.
Acceleration is inversely proportional to mass.
Show Answer & Explanation
Correct Answer: D
Newton's second law states F = ma. Rearranging gives a = F/m, showing that acceleration is inversely proportional to mass when force is constant.
Q4
Medium
A 5 kg object is pulled across a horizontal surface with a constant force of 20 N. If the coefficient of kinetic friction between the object and the surface is 0.2, what is the object's acceleration?
A.
2 m/sยฒ
B.
1 m/sยฒ
C.
3 m/sยฒ
D.
4 m/sยฒ
Show Answer & Explanation
Correct Answer: A
To find the acceleration, we first calculate the frictional force (F_friction = ฮผ * m * g = 0.2 * 5 kg * 9.8 m/sยฒ = 9.8 N). The net force (F_net) is the applied force minus friction (F_net = 20 N - 9.8 N = 10.2 N). Therefore, acceleration (a = F_net/m = 10.2 N / 5 kg = 2 m/sยฒ).
Q5
Medium
A car accelerates from rest to a speed of 30 m/s in 10 seconds. What is the distance covered by the car during this time?
A.
150 m
B.
300 m
C.
200 m
D.
100 m
Show Answer & Explanation
Correct Answer: A
Using the formula for distance with constant acceleration (d = vi * t + 0.5 * a * tยฒ). Here, initial speed (vi) is 0, final speed (vf) is 30 m/s, and time (t) is 10 s. First, calculate acceleration (a = (vf - vi) / t = 30 m/s / 10 s = 3 m/sยฒ). Then, distance d = 0 + 0.5 * 3 m/sยฒ * (10 s)ยฒ = 150 m.
Q6
Medium
An object is in equilibrium under the influence of three forces applied at angles of 0ยฐ, 120ยฐ, and 240ยฐ to the x-axis. If the magnitudes of the forces are 10 N each, what is the net force acting on the object?
A.
0 N
B.
5 N
C.
10 N
D.
15 N
Show Answer & Explanation
Correct Answer: A
In equilibrium, the net force must be zero. The components of the three forces (10 N at 0ยฐ, 120ยฐ, and 240ยฐ) can be calculated and summed. Each force's x and y components will balance out due to symmetry. Since the forces are equally spaced and of equal magnitude, they cancel out, resulting in a net force of 0 N.
Q7
Medium
A block of mass 2 kg rests on a frictionless surface. A force of 12 N is applied to the block. What is the acceleration of the block?
A.
4 m/sยฒ
B.
6 m/sยฒ
C.
2 m/sยฒ
D.
3 m/sยฒ
Show Answer & Explanation
Correct Answer: B
According to Newton's second law, F = ma, where F is the net force, m is mass, and a is acceleration. Rearranging gives a = F/m. Substituting the values: a = 12 N / 2 kg = 6 m/sยฒ.
Q8
Hard
A 2 kg block is sliding down a frictionless incline that makes an angle of 30 degrees with the horizontal. What is the acceleration of the block down the incline?
A.
9.8 m/sยฒ
B.
4.9 m/sยฒ
C.
5.0 m/sยฒ
D.
8.5 m/sยฒ
Show Answer & Explanation
Correct Answer: B
The acceleration of the block can be calculated using the formula a = g * sin(ฮธ) where g = 9.8 m/sยฒ and ฮธ = 30 degrees. Thus, a = 9.8 * sin(30) = 9.8 * 0.5 = 4.9 m/sยฒ.
Q9
Hard
A car of mass 1200 kg is moving at a speed of 20 m/s and suddenly applies brakes, coming to a stop in 5 seconds. What is the average force exerted by the brakes on the car?
A.
4800 N
B.
2400 N
C.
600 N
D.
1200 N
Show Answer & Explanation
Correct Answer: A
To calculate the average force, first determine the deceleration using the formula a = (vf - vi) / t. Here, vf = 0 m/s, vi = 20 m/s, and t = 5 s, giving a = (0 - 20) / 5 = -4 m/sยฒ. Then, apply F = m * a, so F = 1200 kg * (-4 m/sยฒ) = -4800 N. The negative sign indicates the force is acting opposite to the motion.
Q10
Hard
A 2 kg object is subjected to a net force of 10 N. What is the acceleration of the object?
A.
2 m/sยฒ
B.
5 m/sยฒ
C.
7 m/sยฒ
D.
10 m/sยฒ
Show Answer & Explanation
Correct Answer: B
According to Newton's second law (F = ma), acceleration (a) can be calculated by rearranging the formula to a = F/m. Here, F = 10 N and m = 2 kg. Thus, a = 10 N / 2 kg = 5 m/sยฒ, making option B the correct choice.
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Dynamics โ AP (Advanced Placement) AP Physics 1 Practice Questions Online
This page contains 149 practice MCQs for the chapter Dynamics in AP (Advanced Placement) AP Physics 1. The questions are organized by difficulty โ 53 easy, 66 medium, 30 hard โ so you can choose the right level for your preparation.
Every question includes a detailed explanation to help you understand the concept, not just memorize answers. Take a timed quiz to simulate exam conditions, or practice at your own pace with no time limit.