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.
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.
College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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}} =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd2932a05-99f3-41c1-989e-7dbca32c3377%2F6ff752f5-5a19-4a8e-8cff-28fc003a0267%2Fw1ls5ph_processed.png&w=3840&q=75)
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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