College Physics
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ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
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![### Problem Statement
A force given by \( F(x) = 0.5 x^3 \hat{i} \) N acts on a 2 kg mass moving on a frictionless surface. The mass moves from \( x = 3 \) m to \( x = 6 \) m.
1. **(a)** How much work is done by the force?
2. **(b)** If the mass has a speed of 5 m/s at \( x = 3 \), what is its speed at \( x = 6 \) m?
---
### Solution Explanation
#### Part (a): Calculating the Work Done
The work done by the force as the mass moves from \( x = 3 \) m to \( x = 6 \) m can be found using the work integral of the force over the distance:
\[ W = \int_{x_1}^{x_2} F(x) \, dx \]
Given:
\[ F(x) = 0.5 x^3 \]
\[ x_1 = 3 \, \text{m} \]
\[ x_2 = 6 \, \text{m} \]
So,
\[ W = \int_{3}^{6} 0.5 x^3 \, dx \]
Evaluating the integral:
\[ W = 0.5 \int_{3}^{6} x^3 \, dx \]
\[ W = 0.5 \left[ \frac{x^4}{4} \right]_{3}^{6} \]
\[ W = 0.5 \left( \frac{6^4}{4} - \frac{3^4}{4} \right) \]
\[ W = 0.5 \left( \frac{1296}{4} - \frac{81}{4} \right) \]
\[ W = 0.5 \left( 324 - 20.25 \right) \]
\[ W = 0.5 \times 303.75 \]
\[ W = 151.875 \, \text{J} \]
Therefore, the work done by the force is \( 151.875 \, \text{J} \).
---
#### Part (b): Calculating the Speed at \( x = 6 \) m
To determine the speed of the mass at \( x = 6 \)](https://content.bartleby.com/qna-images/question/ba4cba5c-1d49-41b8-aac4-ce3ae0c647d3/0db80878-6efc-40d9-b880-1b4913a96664/y7ru70d_processed.png)
Transcribed Image Text:### Problem Statement
A force given by \( F(x) = 0.5 x^3 \hat{i} \) N acts on a 2 kg mass moving on a frictionless surface. The mass moves from \( x = 3 \) m to \( x = 6 \) m.
1. **(a)** How much work is done by the force?
2. **(b)** If the mass has a speed of 5 m/s at \( x = 3 \), what is its speed at \( x = 6 \) m?
---
### Solution Explanation
#### Part (a): Calculating the Work Done
The work done by the force as the mass moves from \( x = 3 \) m to \( x = 6 \) m can be found using the work integral of the force over the distance:
\[ W = \int_{x_1}^{x_2} F(x) \, dx \]
Given:
\[ F(x) = 0.5 x^3 \]
\[ x_1 = 3 \, \text{m} \]
\[ x_2 = 6 \, \text{m} \]
So,
\[ W = \int_{3}^{6} 0.5 x^3 \, dx \]
Evaluating the integral:
\[ W = 0.5 \int_{3}^{6} x^3 \, dx \]
\[ W = 0.5 \left[ \frac{x^4}{4} \right]_{3}^{6} \]
\[ W = 0.5 \left( \frac{6^4}{4} - \frac{3^4}{4} \right) \]
\[ W = 0.5 \left( \frac{1296}{4} - \frac{81}{4} \right) \]
\[ W = 0.5 \left( 324 - 20.25 \right) \]
\[ W = 0.5 \times 303.75 \]
\[ W = 151.875 \, \text{J} \]
Therefore, the work done by the force is \( 151.875 \, \text{J} \).
---
#### Part (b): Calculating the Speed at \( x = 6 \) m
To determine the speed of the mass at \( x = 6 \)
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