A man pulls out a block of mass m from the depth d with help of a light string. If acceleration of the block is `g/2` then the work done by the string on the block will be

Options

A. `-3mgd /2`
B. `+3mgd /2`
C. `-mgd /2`
D. `+mgd /2`

Answer

To solve the problem step by step, we will analyze the forces acting on the block and calculate the work done by the string.

### Step 1: Identify the forces acting on the block
When the block is being pulled up, two main forces act on it:
- The gravitational force (weight) acting downwards: \( F_g = mg \)
- The tension force \( T \) acting upwards due to the string.

### Step 2: Write the equation of motion
According to Newton's second law, the net force acting on the block can be expressed as:
\[ T - mg = ma \]
where \( a \) is the acceleration of the block. Given that the acceleration \( a = \frac{g}{2} \), we can substitute this into the equation:
\[ T - mg = m \left(\frac{g}{2}\right) \]

### Step 3: Solve for tension \( T \)
Rearranging the equation gives:
\[ T = mg + m\left(\frac{g}{2}\right) \]
\[ T = mg + \frac{mg}{2} = mg\left(1 + \frac{1}{2}\right) = mg \cdot \frac{3}{2} \]
Thus, the tension in the string is:
\[ T = \frac{3mg}{2} \]

### Step 4: Calculate the work done by the string
The work done \( W \) by the string on the block is given by the formula:
\[ W = F \cdot d \]
where \( F \) is the force applied (tension in this case) and \( d \) is the displacement. Since the tension and displacement are in the same direction, we can write:
\[ W = T \cdot d \]
Substituting the value of tension we found:
\[ W = \left(\frac{3mg}{2}\right) \cdot d \]
Thus, the work done by the string on the block is:
\[ W = \frac{3mgd}{2} \]

### Final Answer
The work done by the string on the block is \( \frac{3mgd}{2} \).

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