A man pushes a 80 N crate a distance of 5.0 m upward along a frictionless slope that makes an angle of `30^@` with the horizontal. The force he exerts is parallel to the slope. If the speed of the crate increases at a rate of 1.5 then the work done by the man is :

Options

A. `-200 J `
B. 61 J
C. 140 J
D. 260 J

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the given data
- Weight of the crate (W) = 80 N
- Distance moved along the slope (d) = 5.0 m
- Angle of the slope (θ) = 30°
- Acceleration of the crate (a) = 1.5 m/s²

### Step 2: Calculate the mass of the crate
The weight (W) is related to mass (m) by the equation:
\[ W = mg \]
Where \( g \) (acceleration due to gravity) is approximately \( 10 \, \text{m/s}^2 \).

Thus, we can find the mass:
\[ m = \frac{W}{g} = \frac{80 \, \text{N}}{10 \, \text{m/s}^2} = 8 \, \text{kg} \]

### Step 3: Calculate the component of weight acting along the slope
The component of the weight acting down the slope is given by:
\[ W_{\text{down}} = W \sin(\theta) \]
Substituting the values:
\[ W_{\text{down}} = 80 \, \text{N} \cdot \sin(30°) = 80 \, \text{N} \cdot 0.5 = 40 \, \text{N} \]

### Step 4: Apply Newton's second law to find the force exerted by the man
Using Newton's second law:
\[ F - W_{\text{down}} = ma \]
Where \( F \) is the force exerted by the man.

Substituting the known values:
\[ F - 40 \, \text{N} = 8 \, \text{kg} \cdot 1.5 \, \text{m/s}^2 \]
\[ F - 40 \, \text{N} = 12 \, \text{N} \]
\[ F = 12 \, \text{N} + 40 \, \text{N} = 52 \, \text{N} \]

### Step 5: Calculate the work done by the man
The work done (W) is given by the formula:
\[ W = F \cdot d \cdot \cos(\theta) \]
Since the force exerted by the man is parallel to the slope, \( \theta = 0° \) and \( \cos(0°) = 1 \):
\[ W = F \cdot d \]
Substituting the values:
\[ W = 52 \, \text{N} \cdot 5.0 \, \text{m} = 260 \, \text{J} \]

### Final Answer
The work done by the man is **260 Joules**.

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