Two forces each of magnitude `10 N` act simultaneously on a body with their directions inclined to each other at an angle of `120^@`and displaces the body over `10 m` along the bisector of the angle between the two forces. Then the work done by force is

Options

A. 5 J
B. 1 J
C. 50 J
D. 100 J

Explanation

To solve the problem, we need to determine the work done by the two forces acting on the body. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Forces
We have two forces, each of magnitude \( F = 10 \, \text{N} \), acting at an angle of \( 120^\circ \) to each other.

### Step 2: Determine the Angle Between Forces
Since the forces are inclined at \( 120^\circ \), the angle between the direction of the resultant force and the direction of each individual force can be calculated. The angle between each force and the bisector (the direction of displacement) is \( 60^\circ \) (since \( 120^\circ / 2 = 60^\circ \)).

### Step 3: Resolve Forces into Components
Each force can be resolved into its components along the direction of the displacement (along the bisector). The components of each force along the direction of displacement can be calculated using:
- \( F_{\text{along}} = F \cos(60^\circ) \)

Calculating this gives:
\[ F_{\text{along}} = 10 \, \text{N} \cdot \cos(60^\circ) = 10 \, \text{N} \cdot \frac{1}{2} = 5 \, \text{N} \]

### Step 4: Calculate the Net Force
Since there are two forces acting along the direction of displacement, the total force acting along the bisector is:
\[ F_{\text{net}} = 2 \cdot F_{\text{along}} = 2 \cdot 5 \, \text{N} = 10 \, \text{N} \]

### Step 5: Calculate Work Done
The work done by the net force when the body is displaced over a distance \( S = 10 \, \text{m} \) is given by the formula:
\[ W = F_{\text{net}} \cdot S \]
Substituting the values:
\[ W = 10 \, \text{N} \cdot 10 \, \text{m} = 100 \, \text{J} \]

### Final Answer
The work done by the forces is \( 100 \, \text{J} \).

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