Review A box with mass m is launched up a ramp with coefficient of kinetic friction uk that is inclined at angle a (the angle between the slope and horizontal). The box stops after rising a vertical distance h above the bottom of the incline (note that the actual distance travelled is longer!) Part A Use the work-energy theorem to calculate the speed v that the box has at the bottom of the incline. Express your answer in terms of some or all of the variables m, g, h, lk, and a. • View Available Hint(s) ΑΣφ ? K IT Σ Ф Ω |2gh(1+µ̟cot 2gh(1 cot O) V = K 圓

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### Physics Problem: Work-Energy Theorem Application

**Problem Statement:**
A box with mass \( m \) is launched up a ramp with coefficient of kinetic friction \( \mu_k \) that is inclined at angle \( \alpha \) (the angle between the slope and horizontal). The box stops after rising a vertical distance \( h \) above the bottom of the incline (note that the actual distance traveled is longer!).

**Part A:**
Use the work-energy theorem to calculate the speed \( v \) that the box has at the bottom of the incline.

Express your answer in terms of some or all of the variables \( m \), \( g \), \( h \), \( \mu_k \), and \( \alpha \).

**Solution:**
The speed \( v \) of the box at the bottom of the incline is given by the equation:
\[
v = \sqrt{2gh \left( 1 + \mu_k \cot \alpha \right)}
\]

### Explanation
- **Variables:**
  - \( m \): Mass of the box
  - \( g \): Acceleration due to gravity
  - \( h \): Vertical height the box rises
  - \( \mu_k \): Coefficient of kinetic friction
  - \( \alpha \): Incline angle

- **Equation Details:**
  - The term \( 2gh \) represents the gravitational potential energy converted to kinetic energy.
  - The term \( 1 + \mu_k \cot \alpha \) accounts for the contribution of frictional force along the incline.

### Diagram (Not included, but described hypothetically):
The diagram would ideally show:
- A ramp inclined at an angle \( \alpha \).
- A box starting at the bottom and moving up the ramp.
- Indication of vertical distance \( h \) and the actual distance traveled along the ramp.

This setup helps in visualizing the problem, understanding the forces in play, and the work done against those forces.
Transcribed Image Text:### Physics Problem: Work-Energy Theorem Application **Problem Statement:** A box with mass \( m \) is launched up a ramp with coefficient of kinetic friction \( \mu_k \) that is inclined at angle \( \alpha \) (the angle between the slope and horizontal). The box stops after rising a vertical distance \( h \) above the bottom of the incline (note that the actual distance traveled is longer!). **Part A:** Use the work-energy theorem to calculate the speed \( v \) that the box has at the bottom of the incline. Express your answer in terms of some or all of the variables \( m \), \( g \), \( h \), \( \mu_k \), and \( \alpha \). **Solution:** The speed \( v \) of the box at the bottom of the incline is given by the equation: \[ v = \sqrt{2gh \left( 1 + \mu_k \cot \alpha \right)} \] ### Explanation - **Variables:** - \( m \): Mass of the box - \( g \): Acceleration due to gravity - \( h \): Vertical height the box rises - \( \mu_k \): Coefficient of kinetic friction - \( \alpha \): Incline angle - **Equation Details:** - The term \( 2gh \) represents the gravitational potential energy converted to kinetic energy. - The term \( 1 + \mu_k \cot \alpha \) accounts for the contribution of frictional force along the incline. ### Diagram (Not included, but described hypothetically): The diagram would ideally show: - A ramp inclined at an angle \( \alpha \). - A box starting at the bottom and moving up the ramp. - Indication of vertical distance \( h \) and the actual distance traveled along the ramp. This setup helps in visualizing the problem, understanding the forces in play, and the work done against those forces.
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