You set up a ramp as depicted to practice your skateboard skills. The ramp is a wooden board L=10.0 m long and is propped up between the ground level at one end and h=0.5 m at the other end. You approach the ramp from the right (higher ground) at 11 m/s. How fast are you once you get past the ramp (lower ground)? Assume your mass is 100 kg and friction is negligible.

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### Analyzing the Motion on a Ramp

#### Problem Statement:

You set up a ramp as depicted to practice your skateboard skills. The ramp is a wooden board \( L = 10.0 \) meters long and is elevated by \( h = 0.5 \) meters at one end. You approach the ramp from the right (higher ground) at \( 11 \) meters per second (m/s). Calculate how fast you will be traveling once you get past the ramp (lower ground). Assume your mass is \( 100 \) kilograms and friction is negligible.

#### Diagram Explanation:
The diagram depicts a right-angled triangular ramp with:
- Hypotenuse \( L \) (length of the ramp) labeled \( L \)
- Vertical height \( h \) labeled \( h \)
- Base of the triangle on the ground

### Solution:

To solve this problem, we will apply the principles of conservation of energy. We start by recognizing that the total mechanical energy (potential energy + kinetic energy) of the skateboarder remains constant, given the friction is negligible.

#### Initial Energy at the Top:
1. **Kinetic Energy (KE):**
   \[
   \text{KE}_{\text{initial}} = \frac{1}{2} m v^2
   \]
   where:
   \[
   m = 100 \text{ kg}
   \]
   \[
   v = 11 \text{ m/s}
   \]

   \[
   \text{KE}_{\text{initial}} = \frac{1}{2} \times 100 \text{ kg} \times (11 \text{ m/s})^2
   \]
   \[
   \text{KE}_{\text{initial}} = 0.5 \times 100 \times 121
   \]
   \[
   \text{KE}_{\text{initial}} = 6050 \text{ J}
   \]

2. **Potential Energy (PE):**
   Given that the skateboarder is initially at a height \( h = 0.5 \) m:
   \[
   \text{PE}_{\text{initial}} = mgh
   \]
   where:
   \[
   g = 9.81 \text{ m/s}^2
   \]
   \[
   \text{PE}_{\text{initial}} =
Transcribed Image Text:### Analyzing the Motion on a Ramp #### Problem Statement: You set up a ramp as depicted to practice your skateboard skills. The ramp is a wooden board \( L = 10.0 \) meters long and is elevated by \( h = 0.5 \) meters at one end. You approach the ramp from the right (higher ground) at \( 11 \) meters per second (m/s). Calculate how fast you will be traveling once you get past the ramp (lower ground). Assume your mass is \( 100 \) kilograms and friction is negligible. #### Diagram Explanation: The diagram depicts a right-angled triangular ramp with: - Hypotenuse \( L \) (length of the ramp) labeled \( L \) - Vertical height \( h \) labeled \( h \) - Base of the triangle on the ground ### Solution: To solve this problem, we will apply the principles of conservation of energy. We start by recognizing that the total mechanical energy (potential energy + kinetic energy) of the skateboarder remains constant, given the friction is negligible. #### Initial Energy at the Top: 1. **Kinetic Energy (KE):** \[ \text{KE}_{\text{initial}} = \frac{1}{2} m v^2 \] where: \[ m = 100 \text{ kg} \] \[ v = 11 \text{ m/s} \] \[ \text{KE}_{\text{initial}} = \frac{1}{2} \times 100 \text{ kg} \times (11 \text{ m/s})^2 \] \[ \text{KE}_{\text{initial}} = 0.5 \times 100 \times 121 \] \[ \text{KE}_{\text{initial}} = 6050 \text{ J} \] 2. **Potential Energy (PE):** Given that the skateboarder is initially at a height \( h = 0.5 \) m: \[ \text{PE}_{\text{initial}} = mgh \] where: \[ g = 9.81 \text{ m/s}^2 \] \[ \text{PE}_{\text{initial}} =
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