A block of mass m tied to a string is lowered by a distance d, at a constant acceleration of `g//3`. The work done by the string is

Options

A. `(mgd)/(3)`
B. `(-mgd)/(3)`
C. `(2)/(3)mgd`
D. `(-2)/(3)mgd`

Answer

To solve the problem of finding the work done by the string when a block of mass \( m \) is lowered by a distance \( d \) with a constant acceleration of \( \frac{g}{3} \), we can follow these steps:

### Step 1: Identify the forces acting on the block
The forces acting on the block are:
- The gravitational force \( F_g = mg \) acting downward.
- The tension \( T \) in the string acting upward.

### Step 2: Apply Newton's second law
Since the block is accelerating downward with an acceleration \( a = \frac{g}{3} \), we can apply Newton's second law:
\[ F_{\text{net}} = ma \]
This gives us the equation:
\[ mg - T = ma \]
Substituting \( a = \frac{g}{3} \):
\[ mg - T = m \left(\frac{g}{3}\right) \]

### Step 3: Solve for the tension \( T \)
Rearranging the equation to solve for \( T \):
\[ T = mg - m\left(\frac{g}{3}\right) \]
\[ T = mg - \frac{mg}{3} \]
\[ T = \frac{3mg}{3} - \frac{mg}{3} = \frac{2mg}{3} \]

### Step 4: Calculate the work done by the tension in the string
The work done by the tension \( W_T \) is given by the formula:
\[ W_T = T \cdot d \cdot \cos(\theta) \]
Where \( \theta \) is the angle between the force and the displacement. Since the tension acts upward and the displacement is downward, \( \theta = 180^\circ \) and \( \cos(180^\circ) = -1 \).

Substituting the values:
\[ W_T = \left(\frac{2mg}{3}\right) \cdot d \cdot (-1) \]
\[ W_T = -\frac{2mgd}{3} \]

### Final Answer
The work done by the string is:
\[ W_T = -\frac{2}{3} mgd \]