Solve #3: problem b) ### Physics Problem: Calculating Work Done on a Shipping Crate**Problem Statement:**A 10.0-kg shipping crate is pulled 3.00 m to the right along the floor by an applied force \( P = 35.0 \, \text{N} \) at an angle \( \theta = 32.0^\circ \). The coefficient of kinetic friction between the crate and the floor is \( \mu_k = 0.400 \).#### Questions:1. **How much work is done by the applied force?** - **Formula:** \( W = F \cdot d \cdot \cos \theta \) - **Calculation:** \[ W = (35 \, \text{N}) \cdot (3 \, \text{m}) \cdot \cos(32^\circ) = 89.05 \, \text{J} \]2. **How much work is done by the gravitational force on the crate?** - **Since gravity acts vertically and there is no vertical displacement:** - **Work done by gravity:** \[ W = 0 \, \text{J} \]3. **How much work is done by the kinetic friction force?** - **Formula:** \( W = \mu_k \cdot m \cdot g \cdot d \) - **Calculation:** \[ W = (0.4) \cdot (10 \, \text{kg}) \cdot (9.8 \, \text{m/s}^2) \cdot (3 \, \text{m}) \] \[ W = 117.6 \, \text{J} \]This problem illustrates how to calculate work done on an object by different forces, considering both applied force direction and opposing forces like friction, identifying the effects of each on the total work done.

Given Data: The mass of a shipping crate is, m=10 kg The distance traveled by a crate is, d=3 m The…

Answered: 6HOW much work is done by the…

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6HOW much work is done by the gravitational force on the crate?

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Question

Solve #3: problem b)

### Physics Problem: Calculating Work Done on a Shipping Crate

**Problem Statement:**

A 10.0-kg shipping crate is pulled 3.00 m to the right along the floor by an applied force \( P = 35.0 \, \text{N} \) at an angle \( \theta = 32.0^\circ \). The coefficient of kinetic friction between the crate and the floor is \( \mu_k = 0.400 \).

#### Questions:

1. **How much work is done by the applied force?**

- **Formula:** \( W = F \cdot d \cdot \cos \theta \)
- **Calculation:**
\[
W = (35 \, \text{N}) \cdot (3 \, \text{m}) \cdot \cos(32^\circ) = 89.05 \, \text{J}
\]

2. **How much work is done by the gravitational force on the crate?**

- **Since gravity acts vertically and there is no vertical displacement:**
- **Work done by gravity:**
\[
W = 0 \, \text{J}
\]

3. **How much work is done by the kinetic friction force?**

- **Formula:** \( W = \mu_k \cdot m \cdot g \cdot d \)
- **Calculation:**
\[
W = (0.4) \cdot (10 \, \text{kg}) \cdot (9.8 \, \text{m/s}^2) \cdot (3 \, \text{m})
\]
\[
W = 117.6 \, \text{J}
\]

This problem illustrates how to calculate work done on an object by different forces, considering both applied force direction and opposing forces like friction, identifying the effects of each on the total work done.

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