A body is mass `300 kg` is moved through `10 m` along a smooth inclined plane of inclination angle `30^@`. The work done in moving (in joules) is `(g = 9.8 ms^(-2))`.

Options

A. 4900
B. 9800
C. 14700
D. 2450

Explanation

To solve the problem of calculating the work done in moving a body of mass 300 kg along a smooth inclined plane of inclination angle 30 degrees, we can follow these steps:

### Step 1: Identify the forces acting on the body
The only force acting on the body along the incline is the component of the gravitational force acting parallel to the incline. This force can be calculated using the formula:
\[ F = mg \sin(\theta) \]
where:
- \( m = 300 \, \text{kg} \) (mass of the body)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( \theta = 30^\circ \) (angle of inclination)

### Step 2: Calculate the force
Substituting the values into the equation:
\[ F = 300 \times 9.8 \times \sin(30^\circ) \]
Since \( \sin(30^\circ) = \frac{1}{2} \):
\[ F = 300 \times 9.8 \times \frac{1}{2} \]
\[ F = 300 \times 4.9 \]
\[ F = 1470 \, \text{N} \]

### Step 3: Calculate the work done
The work done \( W \) in moving the body along the incline is given by:
\[ W = F \cdot s \]
where \( s = 10 \, \text{m} \) (distance moved along the incline). Thus:
\[ W = 1470 \times 10 \]
\[ W = 14700 \, \text{J} \]

### Final Answer
The work done in moving the body is:
\[ W = 14700 \, \text{J} \]

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