71. ssm A 15-g bullet is fired from a rifle. It takes 2.50 X 10-3 s for the bullet to travel the length of the barrel, and it exits the barrel with a speed of 715 m/s. Assuming that the acceleration of the bullet is constant, find the average net force exerted on the bullet.
71. ssm A 15-g bullet is fired from a rifle. It takes 2.50 X 10-3 s for the bullet to travel the length of the barrel, and it exits the barrel with a speed of 715 m/s. Assuming that the acceleration of the bullet is constant, find the average net force exerted on the bullet.
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Ch.4 #71 . Please see attached image for Physics question . Thank you
![**Problem 71:**
A 15-g bullet is fired from a rifle. It takes \(2.50 \times 10^{-3}\) seconds for the bullet to travel the length of the barrel, and it exits the barrel with a speed of 715 m/s. Assuming that the acceleration of the bullet is constant, find the average net force exerted on the bullet.
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**Solution Approach:**
1. **Given Data:**
- Mass of the bullet (\(m\)) = 15 g = 0.015 kg (convert to kilograms).
- Time (\(t\)) = \(2.50 \times 10^{-3}\) seconds.
- Final velocity (\(v\)) = 715 m/s.
- Initial velocity (\(u\)) = 0 m/s (assuming the bullet starts from rest).
2. **Find Acceleration (\(a\)):**
- Use the formula: \[ v = u + at \]
- Rearrange to solve for acceleration (\(a\)): \[ a = \frac{v - u}{t} \]
3. **Calculate Force (\(F\)):**
- Use Newton's second law: \[ F = ma \]
Calculate \(a\) and subsequently \(F\) using the given values. This will provide the average net force exerted on the bullet.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb2c8c9d8-cb7d-4bb3-b241-ba372e4ba26e%2Fb536c260-e1ab-47ee-9f2b-6c087f0e74ab%2Fwkwkuu1n_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 71:**
A 15-g bullet is fired from a rifle. It takes \(2.50 \times 10^{-3}\) seconds for the bullet to travel the length of the barrel, and it exits the barrel with a speed of 715 m/s. Assuming that the acceleration of the bullet is constant, find the average net force exerted on the bullet.
---
**Solution Approach:**
1. **Given Data:**
- Mass of the bullet (\(m\)) = 15 g = 0.015 kg (convert to kilograms).
- Time (\(t\)) = \(2.50 \times 10^{-3}\) seconds.
- Final velocity (\(v\)) = 715 m/s.
- Initial velocity (\(u\)) = 0 m/s (assuming the bullet starts from rest).
2. **Find Acceleration (\(a\)):**
- Use the formula: \[ v = u + at \]
- Rearrange to solve for acceleration (\(a\)): \[ a = \frac{v - u}{t} \]
3. **Calculate Force (\(F\)):**
- Use Newton's second law: \[ F = ma \]
Calculate \(a\) and subsequently \(F\) using the given values. This will provide the average net force exerted on the bullet.
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