***Physics Work Calculation Problem*****Problem:**A rope exerts a force \( F \) on a 20 kg crate. The crate starts from rest and accelerates upward at 5.00 m/s² near the surface of Earth. How much work was done by the force \( F \) in raising the crate 4.0 m above the floor?**Solution:**To solve this problem, we need to consider the following equations and concepts:1. **Newton’s Second Law**: \[ F_{\text{net}} = ma \] Where: - \( F_{\text{net}} \) is the net force acting on the crate. - \( m \) is the mass of the crate (20 kg). - \( a \) is the acceleration (5.00 m/s²).2. **Forces acting on the crate**: - The gravitational force (\( F_g \)) acting downward. - The applied force (\( F \)) acting upward. The gravitational force is given by: \[ F_g = mg \] Where: - \( g \) is the acceleration due to gravity (9.8 m/s²).3. **Net Force**: \[ F_{\text{net}} = F - F_g \]4. **Calculating the net force**: \[ F_{\text{net}} = ma \] Thus, \[ F - mg = ma \] \[ F = m(a + g) \] \[ F = 20\, \text{kg} (5.00\, \text{m/s}^2 + 9.8\, \text{m/s}^2) \] \[ F = 20\, \text{kg} (14.8\, \text{m/s}^2) \] \[ F = 296\, \text{N} \] 5. **Work Done**: Work done by the force \( F \) in raising the crate is calculated using the formula: \[ W = F \cdot d \] Where \( d \) is the distance (4.0 m). \[ W = 296\, \text{N} \cdot 4.0\, \text{m} \] \[ W =

Mass= 20kg Initial velocity= 0 m/s Acceleration= 5m/s2

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A rope exerts a force F on a 2O.kg crate. The crate Stats in raising the crate 4.0m above the floor?

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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Question

***Physics Work Calculation Problem***

**Problem:**

A rope exerts a force \( F \) on a 20 kg crate. The crate starts from rest and accelerates upward at 5.00 m/s² near the surface of Earth. How much work was done by the force \( F \) in raising the crate 4.0 m above the floor?

**Solution:**

To solve this problem, we need to consider the following equations and concepts:

1. **Newton’s Second Law**:
\[ F_{\text{net}} = ma \]

Where:
- \( F_{\text{net}} \) is the net force acting on the crate.
- \( m \) is the mass of the crate (20 kg).
- \( a \) is the acceleration (5.00 m/s²).

2. **Forces acting on the crate**:
- The gravitational force (\( F_g \)) acting downward.
- The applied force (\( F \)) acting upward.

The gravitational force is given by:
\[ F_g = mg \]
Where:
- \( g \) is the acceleration due to gravity (9.8 m/s²).

3. **Net Force**:
\[ F_{\text{net}} = F - F_g \]

4. **Calculating the net force**:
\[ F_{\text{net}} = ma \]
Thus,
\[ F - mg = ma \]
\[ F = m(a + g) \]
\[ F = 20\, \text{kg} (5.00\, \text{m/s}^2 + 9.8\, \text{m/s}^2) \]
\[ F = 20\, \text{kg} (14.8\, \text{m/s}^2) \]
\[ F = 296\, \text{N} \]

5. **Work Done**:
Work done by the force \( F \) in raising the crate is calculated using the formula:
\[ W = F \cdot d \]
Where \( d \) is the distance (4.0 m).

\[ W = 296\, \text{N} \cdot 4.0\, \text{m} \]
\[ W =

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