A block of mass m=1/3 kg is kept on a rough horizontal plane. Friction coefficient is `mu=0.75`. The work done by minimum force required to drag to the block along the plane by a distance 5m is :-

Options

A. 8 J
B. 4 J
C. 2 J
D. 6 J

Answer

To solve the problem, we need to find the work done by the minimum force required to drag a block of mass \( m = \frac{1}{3} \) kg on a rough horizontal plane with a coefficient of friction \( \mu = 0.75 \) over a distance of 5 m.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block:**
- The weight of the block \( W = mg \) acts downwards.
- The normal force \( R \) acts upwards.
- The frictional force \( F_f \) acts opposite to the direction of motion, given by \( F_f = \mu R \).
- The applied force \( P \) is what we need to determine.

2. **Calculate the Weight of the Block:**
\[ W = mg = \frac{1}{3} \text{ kg} \times 10 \text{ m/s}^2 = \frac{10}{3} \text{ N} \]

3. **Set Up the Equation for Normal Force:**
Since the block is on a horizontal plane, the normal force \( R \) can be expressed as:
\[ R = W = \frac{10}{3} \text{ N} \]

4. **Calculate the Frictional Force:**
The frictional force is given by:
\[ F_f = \mu R = 0.75 \times \frac{10}{3} = \frac{7.5}{3} = 2.5 \text{ N} \]

5. **Determine the Minimum Force Required to Overcome Friction:**
The minimum force \( P \) required to drag the block is equal to the frictional force when the block is just about to move:
\[ P = F_f = 2.5 \text{ N} \]

6. **Calculate the Work Done by the Force:**
Work done \( W \) by the force \( P \) when moving the block a distance \( d = 5 \) m is given by:
\[ W = P \times d = 2.5 \text{ N} \times 5 \text{ m} = 12.5 \text{ J} \]

### Final Answer:
The work done by the minimum force required to drag the block along the plane by a distance of 5 m is \( 12.5 \) Joules.

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