2. A cardboard box with a mass of 10.0 kg is at rest on a ramp with an angle of 25°. a. Determine the coefficient of static friction between the box and the ramp. P = 0.466 FN=mg cos(25) F = NFN msin (25) = Umg cos (25) mgcos(25) mgcos (25) mgcos(25 3. A box of mass m = 10.25 kg is at rest on a rough floor. The coefficians f 568 b. A 68.0 N force is applied parallel to the ramp causing the box to move up the incline at a constant velocity. Calculate the coefficient of kinetic friction between the box and the ramp. a=0m/s² mgcos25 ma masin 25 mg FN FN mgsin 25 250 m = 10

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# AP Physics 1 Review Questions

## Integrating Content, Inquiry, and Reasoning

### 1. Dynamics on a Frictional Surface

A box of mass \( m_1 = 2.5 \, \text{kg} \) rests atop a box of mass \( m_2 = 3.0 \, \text{kg} \). Box \( m_2 \) rests on a frictionless surface. Box \( m_1 \) is pulled by a 95 N force and slides with friction (\( \mu_k = 0.33 \)) across the top of box \( m_2 \). What is the acceleration of box \( m_2 \)?

- **Diagram:** 
  - Box \( m_1 \) is shown with a force arrow pointing left labeled \( 95 \, \text{N} \).
  - Frictional force \( f = \mu X N = 0.33 \times (24.5) = 8.09 \, \text{N} \).
  - Calculations show \( a_{m2} = 2.70 \, \text{m/s}^2 \).

### 2. Static Friction on a Ramp

A cardboard box with a mass of 10.0 kg is at rest on a ramp with an angle of 25°.

#### a. Determine the coefficient of static friction between the box and the ramp.

- Calculations show \( \mu = 0.466 \).
- Diagram:
  - Forces at an angle of 25° imply \( F_N = mg \cos(25) \).
  - \( mg \sin(25) = \mu mg \cos(25) \).

#### b. A 68.0 N force is applied parallel to the ramp, causing the box to move up the incline at a constant velocity. Calculate the coefficient of kinetic friction between the box and the ramp.

- Calculation of acceleration \( a = 0 \, \text{m/s}^2 \).

### 3. Friction on a Rough Floor

A box of mass \( m = 10.25 \, \text{kg} \) is at rest on a rough floor. The coefficient of static friction between the box and the floor is \( \mu_s = 0.35 \). A rope is attached to the box and pulled at an upward angle of
Transcribed Image Text:# AP Physics 1 Review Questions ## Integrating Content, Inquiry, and Reasoning ### 1. Dynamics on a Frictional Surface A box of mass \( m_1 = 2.5 \, \text{kg} \) rests atop a box of mass \( m_2 = 3.0 \, \text{kg} \). Box \( m_2 \) rests on a frictionless surface. Box \( m_1 \) is pulled by a 95 N force and slides with friction (\( \mu_k = 0.33 \)) across the top of box \( m_2 \). What is the acceleration of box \( m_2 \)? - **Diagram:** - Box \( m_1 \) is shown with a force arrow pointing left labeled \( 95 \, \text{N} \). - Frictional force \( f = \mu X N = 0.33 \times (24.5) = 8.09 \, \text{N} \). - Calculations show \( a_{m2} = 2.70 \, \text{m/s}^2 \). ### 2. Static Friction on a Ramp A cardboard box with a mass of 10.0 kg is at rest on a ramp with an angle of 25°. #### a. Determine the coefficient of static friction between the box and the ramp. - Calculations show \( \mu = 0.466 \). - Diagram: - Forces at an angle of 25° imply \( F_N = mg \cos(25) \). - \( mg \sin(25) = \mu mg \cos(25) \). #### b. A 68.0 N force is applied parallel to the ramp, causing the box to move up the incline at a constant velocity. Calculate the coefficient of kinetic friction between the box and the ramp. - Calculation of acceleration \( a = 0 \, \text{m/s}^2 \). ### 3. Friction on a Rough Floor A box of mass \( m = 10.25 \, \text{kg} \) is at rest on a rough floor. The coefficient of static friction between the box and the floor is \( \mu_s = 0.35 \). A rope is attached to the box and pulled at an upward angle of
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