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**STOP** β Before you look at the answers make sure you gave this practice quiz a try so you can assess your understanding of the concepts covered in Unit 2. Click here for the practice questions:

**AP Physics 1 Unit 2 Multiple Choice Questions**.

**Facts about the test**: The AP Physics 1 exam has 50 multiple choice questions (45 single-select and 5 multiple-select) and you will be given 90 minutes to complete the section. That means it should take you around 15 minutes to complete 8 questions.

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The following questions were not written by College Board and, although they cover information outlined in the **AP Physics 1 Course and Exam Description**, the formatting on the exam may be different.*

**1.Β A 50.0 N box is at rest on a horizontal surface. The coefficient of static friction between the box and the surface is 0.50, and the coefficient of kinetic friction is 0.40. A horizontal 20.0 N force is then exerted on the box. The magnitude of the acceleration of the box is most nearly**

**A. 0 m/s/s**

B. 1 m/s/s

C. 2 m/s/s

D.4 m/s/s

**Answer**: The frictional force on the box can be found using f = mu * Fn, which is 25N (use the static coefficient because the box isn't moving). This is greater than the applied force of 20N, so the box will remain stationary

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**Study AP Physics 1, Unit 2.3: ****Contact Forces**

**2.Β The boxes are released from rest, assuming friction is negligible what is the acceleration of the 2kg block?**

A. 0 m/s/s

**B. 3.3 m/s/s**

C. 5 m/s/s

D. 10 m/s/s

**Answer**: Use Newton's 2nd Law and treat both boxes as a single system. The weight of the 2kg block is the only outside force acting on the system. So F = ma => 20N = 6a => a = 20/6 = 3.3 m/s/s

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**Study AP Physics 1, Unit 2.6: ****Newton's Second Law**

**3. If the boxes are held in equilibrium, what is the minimum coefficient of static friction between the 4.0kg block and the tabletop?**

A. 0.25

**B. 0.5**

C. 0.75

D. 1

**Answer**: If the blocks are in equilibrium, then the weight of the 2.0kg block must be equal to the frictional force between the 4.0kg block and the table. 20N = mu(Fn) => 20N = mu(40N) => mu = 0.5

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**Study AP Physics 1, Unit 2.4: ****Newton's First Law**

**4.Β Each of these graphs show the motion of an object traveling in a straight line.**

**Which of these graphs shows an object in equilibrium?**

A. I

B. II

C. III

**D. I and II**

**
Answer**: In equilibrium, the object's acceleration is zero. Therefore the velocity must be constant. Since the slope of a distance vs time graph is velocity, I and II are correct

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**Study AP Physics 1, Unit 2.4: ****Newton's First Law**

**5.Β Each of these graphs show the motion of an object traveling in a straight line.**

**Which of these graphs shows an object with a non-zero net force?**

A. I

B. II

**C. III**

D. I and II

**Answer**: A non-zero net force means that the object must be accelerating, so the slope of a distance vs time graph must be changing. Only graph III shows that.

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**Study AP Physics 1, Unit 2.6:** **Newton's Second Law**

**6. The weight of an 10kg box is approximately**

A. 1N

B. 10N

**C. 100N**

D. 1000N

**Answer**: Fg = mg so Fg = 10kg(10m/s/s) = 100N

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**Study AP Physics 1, Unit 2.7:** **Applications of Newton's Second Law**

**7.Β What is the acceleration of this object?**

A. 0 m/s/s

**B. 1 m/s/s**

C. 2 m/s/s

D. 3 m/s/s

**Answer**: Newton's 2nd Law: The net force on the object is 10N (20-10) and the mass is 10kg, so F = ma => a = 10/10 = 1m/s/s

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**Study AP Physics 1, Unit 2.2: ****The Gravitational Field**

**8.Β What is the magnitude of the normal force acting on the box?**

A. 0N

B. 10kg

**C. 100N**

D. Can't be determined

**Answer**: The normal force is caused by the table pushing up on the box. It is an action-reaction forces, which according to Newton's 3rd Law must be equal to the weight of the box (100N)

**9.Β **

A. 30N

**B. 50N**

C. 100N

D. 150N

**Answer**: Calculate the force of static friction using f = mu * Fn where Fn = mg = 100N. Use static friction because the box is at rest.

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**Study AP Physics 1, Unit 2.3: ****Contact Forces**

**10.Β **

A. 0.5 m/s/s

B. 1 m/s/s

**C. 5 m/s/s**

D. 10 m/s/s

**Answer**: The net force on the box is the parallel component of the weight (mgsin(Q) = 100sin(30) = 50N). Use Newton's 2nd Law to calculate the acceleration (F = ma => 50N = 10kg (a) => a = 5m/s/s)

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**Study AP Physics 1, Unit 2.6: ****Newton's Second Law**

**11.Β **

A. 0N

B. 50N

**C. 87N**

D. 100N

**Answer**: The normal force must be equal to the perpendicular component of the weight. (Fn = mgcos(Q) = 100 cos(30) = 87N)

**12.Β A person stands on a scale in an elevator. If the scale reads 600 N when that person is riding upward at a constant velocity of 10 m/s, what is the scale reading when the elevator is at rest?**

A. 200N

B. 400N

**C. 600N**

D. 800N

**Answer**: Constant velocity means that the forces on the person are balanced (Fscale = Fg). This also is true when the person is stationary (constant 0 velocity). The only times when the scale will not be the same as the weight is when the elevator is accelerating.

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**Study AP Physics 1, Unit 2.7: ****Applications of Newton's Second Law**

**13.Β A student records the following velocity vs time data watching a 0.2kg block slide across a rough horizontal surface.Β **

**What is the magnitude of the frictional force acting on the block?**

A. 0.08N

**B. 0.16N**

C. 0.799N

D. 2N

**Answer**: The slope of the graph is the acceleration. Since the only unbalanced force acting on the block is friction, use F = ma to get 0.16N

**14.Β A simple Atwood machine is created by hanging two blocks over a fixed, massless pulley. If the blocks are released, what is the acceleration of the 5kg block?**

**A. 1.67 m/s/s**

B. 2.86 m/s/s

C. 4 m/s/s

D. 14 m/s/s

**Answer**: Use Newton's 2nd Law and treat both boxes as a single system. F = ma => 70N - 50N = (7kg + 5kg) a => a = 20N / 12kg = 1.67m/s/s

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**Study AP Physics 1, Unit 2.6:** **Newton's Second Law**

**15.Β A cart (m1 = 2kg) is setup so it slides across the tabletop. Itβs attached to a hanging mass (m2 = 10kg) and released from rest. The coefficient of kinetic friction between the cart and the tabletop is 0.5. What is the acceleration of the cart?**

**A. 7.5 m/s/s**

B. 8.25 m/s/s

C. 9.17 m/s/s

D. 10 m/s/s

**Answer**: Use Newton's 2nd Law and treat the cart and the hanging mass as a single system. F = ma => M2g - Ff = (M1+M2)a => 100N - 0.5(20N) = 12kg (a) => a = 90N / 12kg = 7.5m/s/s

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**Study AP Physics 1, Unit 2.6: ****Newton's Second Law**

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