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. `2/3mgh`
B. `(-mgh)/(3)`
C. `mgh`
D. `4/3mgh`

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 weight of the block, \( W = mg \) (acting downward).
- The tension in the string, \( T \) (acting upward).

### Step 2: Apply Newton's Second Law
Since the block is accelerating downward with an acceleration of \( \frac{g}{3} \), we can apply Newton's second law:
\[ F_{\text{net}} = ma \]
Where \( a = \frac{g}{3} \). The net force acting on the block can be expressed as:
\[ mg - T = ma \]
Substituting \( a \) into the equation gives:
\[ mg - T = m \left(\frac{g}{3}\right) \]

### Step 3: Solve for 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 String
The work done by the string is given by the formula:
\[ W = F \cdot d \cdot \cos(\theta) \]
Here, \( F \) is the tension \( T \), \( d \) is the distance moved (which is \( d \)), and \( \theta \) is the angle between the force and the direction of displacement. Since the tension acts upward and the displacement is downward, the angle \( \theta = 180^\circ \) (cosine of 180 degrees is -1).

Substituting the values:
\[ W = T \cdot d \cdot \cos(180^\circ) \]
\[ W = \left(\frac{2mg}{3}\right) \cdot d \cdot (-1) \]
\[ W = -\frac{2mgd}{3} \]

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

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