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 solve the problem of finding the work done by the tension in the string while pulling a block of mass \( m \) vertically upward 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 tension \( T \) in the string acting upwards.
- The gravitational force \( mg \) acting downwards.

### Step 2: Apply Newton's Second Law
Since the block is moving upwards with an acceleration \( a = \frac{g}{3} \), we can apply Newton's second law:

\[ \text{Net force} = \text{mass} \times \text{acceleration} \]

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:

\[ T = mg + m \left(\frac{g}{3}\right) \]

\[ T = mg + \frac{mg}{3} = mg \left(1 + \frac{1}{3}\right) = mg \left(\frac{4}{3}\right) \]

Thus, the tension \( T \) in the string is:

\[ T = \frac{4mg}{3} \]

### Step 4: Calculate the Work Done by the Tension
The work done by the tension \( W \) when moving the block a distance \( h \) is given by the formula:

\[ W = T \cdot h \]

Substituting the expression for \( T \):

\[ W = \left(\frac{4mg}{3}\right) \cdot h = \frac{4mgh}{3} \]

### Final Answer
The work done by the tension in the string is:

\[ \boxed{\frac{4mgh}{3}} \]

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