A child and a sled with a combined mass of 50.0 kg slide down a frictionless slope. If the sled starts from rest and has a speed of 3.00 m/ s at the bottom, what is the height of the hill?
A child and a sled with a combined mass of 50.0 kg slide down a frictionless slope. If the sled starts from rest and has a speed of 3.00 m/ s at the bottom, what is the height of the hill?
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I am getting stuck at the arithmetic part. Please let me know the steps in detail. TIA
![### Example Physics Problem: Kinetic and Potential Energy
**Problem P.5.33**
A child and a sled with a combined mass of 50.0 kg slide down a frictionless slope. If the sled starts from rest and has a speed of 3.00 m/s at the bottom, what is the height of the hill?
---
**Explanation:**
Given data:
- Combined mass (m) = 50.0 kg
- Final speed (v) = 3.00 m/s
- Initial speed (u) = 0 m/s (starts from rest)
- Acceleration due to gravity (g) = 9.8 m/s² (standard value)
To find: Height of the hill (h)
### Solution:
Using the principle of conservation of energy, the potential energy lost by the sled-child system as it descends the hill is converted to kinetic energy.
At the top of the hill (initial position):
- Kinetic Energy (KE_initial) = 0 (because the speed is 0)
- Potential Energy (PE_initial) = mgh
At the bottom of the hill (final position):
- Kinetic Energy (KE_final) = 0.5 * m * v²
- Potential Energy (PE_final) = 0 (since the height is 0)
According to the conservation of energy:
\[ PE_{\text{initial}} = KE_{\text{final}} \]
So,
\[ mgh = \frac{1}{2} mv^2 \]
Simplifying,
\[ gh = \frac{1}{2} v^2 \]
Therefore, solving for h:
\[ h = \frac{v^2}{2g} \]
Substituting the known values:
\[ h = \frac{(3.00 \, \text{m/s})^2}{2 \times 9.8 \, \text{m/s}^2} \]
\[ h = \frac{9.00 \, \text{m}^2/\text{s}^2}{19.6 \, \text{m/s}^2} \]
\[ h = 0.459 \, \text{m} \]
So, the height of the hill is approximately 0.459 meters.
This example demonstrates the application of the conservation of energy principle in physics to determine the height of a hill based on the mass and final velocity of](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fffcb3307-3357-4e27-9ea6-47d78c633263%2F2e0cd2d8-f1d5-41fa-882e-2578e38bdaa1%2F2dqouo.jpeg&w=3840&q=75)
Transcribed Image Text:### Example Physics Problem: Kinetic and Potential Energy
**Problem P.5.33**
A child and a sled with a combined mass of 50.0 kg slide down a frictionless slope. If the sled starts from rest and has a speed of 3.00 m/s at the bottom, what is the height of the hill?
---
**Explanation:**
Given data:
- Combined mass (m) = 50.0 kg
- Final speed (v) = 3.00 m/s
- Initial speed (u) = 0 m/s (starts from rest)
- Acceleration due to gravity (g) = 9.8 m/s² (standard value)
To find: Height of the hill (h)
### Solution:
Using the principle of conservation of energy, the potential energy lost by the sled-child system as it descends the hill is converted to kinetic energy.
At the top of the hill (initial position):
- Kinetic Energy (KE_initial) = 0 (because the speed is 0)
- Potential Energy (PE_initial) = mgh
At the bottom of the hill (final position):
- Kinetic Energy (KE_final) = 0.5 * m * v²
- Potential Energy (PE_final) = 0 (since the height is 0)
According to the conservation of energy:
\[ PE_{\text{initial}} = KE_{\text{final}} \]
So,
\[ mgh = \frac{1}{2} mv^2 \]
Simplifying,
\[ gh = \frac{1}{2} v^2 \]
Therefore, solving for h:
\[ h = \frac{v^2}{2g} \]
Substituting the known values:
\[ h = \frac{(3.00 \, \text{m/s})^2}{2 \times 9.8 \, \text{m/s}^2} \]
\[ h = \frac{9.00 \, \text{m}^2/\text{s}^2}{19.6 \, \text{m/s}^2} \]
\[ h = 0.459 \, \text{m} \]
So, the height of the hill is approximately 0.459 meters.
This example demonstrates the application of the conservation of energy principle in physics to determine the height of a hill based on the mass and final velocity of
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