Bob pushes on a frictionless cart of mass 6.8 kg and the cart has an acceleration of 5.7 m/s2. If Bob now pushes twice as hard on a cart with twice the mass, what will be the acceleration of the heavier cart? Your Answer:
Bob pushes on a frictionless cart of mass 6.8 kg and the cart has an acceleration of 5.7 m/s2. If Bob now pushes twice as hard on a cart with twice the mass, what will be the acceleration of the heavier cart? Your Answer:
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
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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![**Problem:**
Bob pushes on a frictionless cart of mass 6.8 kg, and the cart has an acceleration of 5.7 m/s². If Bob now pushes twice as hard on a cart with twice the mass, what will be the acceleration of the heavier cart?
**Solution:**
To find the acceleration of the heavier cart when Bob pushes twice as hard, we follow these steps:
Given:
- Initial mass of the cart (m₁) = 6.8 kg
- Acceleration of the cart (a₁) = 5.7 m/s²
- Force exerted by Bob (F) = m₁ * a₁
We need to determine:
- The acceleration (a₂) of the new cart with twice the mass when Bob pushes with twice the force.
Step-by-step calculation:
1. Calculate the force Bob initially exerts:
\[ F = m₁ \cdot a₁ = 6.8 \text{ kg} \cdot 5.7 \text{ m/s}^2 = 38.76 \text{ N} \]
2. If Bob pushes twice as hard, the force will be:
\[ F_{\text{new}} = 2 \cdot F = 2 \cdot 38.76 \text{ N} = 77.52 \text{ N} \]
3. The new cart has twice the mass:
\[ m₂ = 2 \cdot m₁ = 2 \cdot 6.8 \text{ kg} = 13.6 \text{ kg} \]
4. Using Newton's second law, calculate the acceleration (a₂) of the heavier cart:
\[ a₂ = \frac{F_{\text{new}}}{m₂} = \frac{77.52 \text{ N}}{13.6 \text{ kg}} \]
5. Compute the result:
\[ a₂ = 5.7 \text{ m/s}^2 \]
Thus, the acceleration of the heavier cart will be \( 5.7 \text{ m/s}^2 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd99dc5ee-0347-4b7c-8511-9a91331fbdb4%2Ff233d209-2c20-4702-b35b-38bc5edf2486%2Fia95au_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem:**
Bob pushes on a frictionless cart of mass 6.8 kg, and the cart has an acceleration of 5.7 m/s². If Bob now pushes twice as hard on a cart with twice the mass, what will be the acceleration of the heavier cart?
**Solution:**
To find the acceleration of the heavier cart when Bob pushes twice as hard, we follow these steps:
Given:
- Initial mass of the cart (m₁) = 6.8 kg
- Acceleration of the cart (a₁) = 5.7 m/s²
- Force exerted by Bob (F) = m₁ * a₁
We need to determine:
- The acceleration (a₂) of the new cart with twice the mass when Bob pushes with twice the force.
Step-by-step calculation:
1. Calculate the force Bob initially exerts:
\[ F = m₁ \cdot a₁ = 6.8 \text{ kg} \cdot 5.7 \text{ m/s}^2 = 38.76 \text{ N} \]
2. If Bob pushes twice as hard, the force will be:
\[ F_{\text{new}} = 2 \cdot F = 2 \cdot 38.76 \text{ N} = 77.52 \text{ N} \]
3. The new cart has twice the mass:
\[ m₂ = 2 \cdot m₁ = 2 \cdot 6.8 \text{ kg} = 13.6 \text{ kg} \]
4. Using Newton's second law, calculate the acceleration (a₂) of the heavier cart:
\[ a₂ = \frac{F_{\text{new}}}{m₂} = \frac{77.52 \text{ N}}{13.6 \text{ kg}} \]
5. Compute the result:
\[ a₂ = 5.7 \text{ m/s}^2 \]
Thus, the acceleration of the heavier cart will be \( 5.7 \text{ m/s}^2 \).
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