A body of mass `5 kg` is moved up over `10 m` along the line of greatest slope of a smooth inclined plane of inclination `30^@` with the horizontal. If `g = 10 m//s^(2)`, the work done will be

Options

A. 500 J
B. 2500 J
C. 250 J
D. 25 J

Explanation

To solve the problem, we need to calculate the work done when a body of mass \(5 \, \text{kg}\) is moved up a smooth inclined plane of inclination \(30^\circ\) over a distance of \(10 \, \text{m}\). The acceleration due to gravity \(g\) is given as \(10 \, \text{m/s}^2\).

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Body:**
The weight of the body acts downwards and can be calculated using the formula:
\[ W = mg \]
where \(m = 5 \, \text{kg}\) and \(g = 10 \, \text{m/s}^2\).

2. **Calculate the Weight of the Body:**
\[ W = 5 \, \text{kg} \times 10 \, \text{m/s}^2 = 50 \, \text{N} \]

3. **Determine the Component of Weight Acting Along the Incline:**
The component of the weight acting down the incline can be found using:
\[ F_{\text{parallel}} = W \sin(\theta) \]
where \(\theta = 30^\circ\).

4. **Calculate \(F_{\text{parallel}}\):**
\[ F_{\text{parallel}} = 50 \, \text{N} \times \sin(30^\circ) = 50 \, \text{N} \times \frac{1}{2} = 25 \, \text{N} \]

5. **Calculate the Work Done Against This Force:**
Work done \(W_d\) is given by the formula:
\[ W_d = F_{\text{parallel}} \times d \]
where \(d = 10 \, \text{m}\).

6. **Substitute the Values:**
\[ W_d = 25 \, \text{N} \times 10 \, \text{m} = 250 \, \text{J} \]

### Final Answer:
The work done in moving the body up the incline is \(250 \, \text{J}\).

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