A block of mass 2 kg is hanging over a smooth and light pulley through a light string. The other end of the string is pulled by a constant force F = 40 N. The kinetic energy of the particle increases 40 J in a given interval of time. Then, `(g = 10 m//s^(2))`

Options

A. Tension in the string is 40 N
B. Displacement of the block in the given interval of time is 2m
C. Workdone by gravity is –20 J
D. Workdone by tension is 80 J

Explanation

To solve the problem step by step, we will analyze the forces acting on the block and apply the work-energy principle.

### Step 1: Identify the forces acting on the block
The forces acting on the block of mass \( m = 2 \, \text{kg} \) are:
1. The weight of the block acting downwards:
\[ W = m \cdot g = 2 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 20 \, \text{N} \]
2. The tension \( T \) in the string acting upwards.

The problem states that a constant force \( F = 40 \, \text{N} \) is pulling the other end of the string.

### Step 2: Apply the Work-Energy Principle
According to the work-energy principle, the net work done on the block is equal to the change in kinetic energy. We know that the kinetic energy increases by \( 40 \, \text{J} \).

The net work done \( W_{\text{net}} \) can be expressed as:
\[ W_{\text{net}} = W_{\text{tension}} + W_{\text{gravity}} \]
Where:
- \( W_{\text{tension}} \) is the work done by the tension.
- \( W_{\text{gravity}} \) is the work done by gravity.

### Step 3: Calculate the work done by gravity
The work done by gravity when the block moves a distance \( s \) downwards is given by:
\[ W_{\text{gravity}} = -W \cdot s = -20 \, \text{N} \cdot s \]
Since we need to find the displacement \( s \), we will calculate it later.

### Step 4: Calculate the work done by tension
The work done by tension when the block moves a distance \( s \) upwards is given by:
\[ W_{\text{tension}} = T \cdot s = T \cdot s \]
Where \( T \) is the tension in the string.

### Step 5: Set up the equation
From the work-energy principle:
\[ W_{\text{net}} = W_{\text{tension}} + W_{\text{gravity}} = 40 \, \text{J} \]
Substituting the expressions for work:
\[ T \cdot s - 20 \cdot s = 40 \]
This simplifies to:
\[ (T - 20) \cdot s = 40 \]

### Step 6: Determine the tension in the string
Since the force \( F = 40 \, \text{N} \) is pulling the string, and the system is in equilibrium, the tension \( T \) in the string must also be equal to \( 40 \, \text{N} \).

### Step 7: Substitute the tension value
Substituting \( T = 40 \, \text{N} \) into the equation:
\[ (40 - 20) \cdot s = 40 \]
This simplifies to:
\[ 20 \cdot s = 40 \]
Thus,
\[ s = \frac{40}{20} = 2 \, \text{m} \]

### Step 8: Calculate work done by tension
Now, we can calculate the work done by tension:
\[ W_{\text{tension}} = T \cdot s = 40 \, \text{N} \cdot 2 \, \text{m} = 80 \, \text{J} \]

### Final Results
1. The displacement of the block \( s = 2 \, \text{m} \).
2. The work done by tension \( W_{\text{tension}} = 80 \, \text{J} \).
3. The work done by gravity \( W_{\text{gravity}} = -20 \cdot 2 = -40 \, \text{J} \).

### Summary
- Tension in the string: \( 40 \, \text{N} \)
- Displacement of the block: \( 2 \, \text{m} \)
- Work done by gravity: \( -40 \, \text{J} \)
- Work done by tension: \( 80 \, \text{J} \)