A string is used to pull a block of mass m vertically up by a distance h at a constant acceleration `(g)/(3)`. The work done by the tension in the string is

Options

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

Answer

To find the work done by the tension in the string while pulling the block of mass \( m \) vertically up by a distance \( h \) 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 downwards.
- The tension in the string, \( T \), acting upwards.

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

### Step 3: Solve for tension \( T \)
Rearranging the equation gives:
\[ T = mg + m\left(\frac{g}{3}\right) \]
Factoring out \( mg \):
\[ T = mg \left(1 + \frac{1}{3}\right) = mg \left(\frac{4}{3}\right) \]
Thus, the tension in the string is:
\[ T = \frac{4}{3}mg \]

### Step 4: Calculate the work done by the tension
The work done by a constant force is given by:
\[ W = F \cdot d \cdot \cos(\theta) \]
where \( F \) is the force, \( d \) is the displacement, and \( \theta \) is the angle between the force and the displacement. In this case, the tension and the displacement are in the same direction, so \( \theta = 0^\circ \) and \( \cos(0) = 1 \).

Thus, the work done by the tension is:
\[ W = T \cdot h \cdot \cos(0) = T \cdot h \]
Substituting the value of \( T \):
\[ W = \left(\frac{4}{3}mg\right) \cdot h \]
Therefore, the work done by the tension in the string is:
\[ W = \frac{4}{3}mgh \]

### Final Answer
The work done by the tension in the string is \( \frac{4}{3}mgh \).

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