A block of mass M 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. The kinetic energy of the block increases by 20 J in 1 s:-

Options

A. The tension in the string is Mg.
B. The tension in the string is F
C. The work done by the tension on the block is 20 J in the above 1 s.
D. The work done by the force of gravity is 20J in the above 1 s.

Explanation

To solve the problem, we need to analyze the forces acting on the block and how they relate to the change in kinetic energy.

### Step 1: Understand the System
We have a block of mass \( M \) hanging from a pulley, and the other end of the string is being pulled by a constant force \( F \). The system is in motion, and we know that the kinetic energy of the block increases by 20 J in 1 second.

### Step 2: Identify Forces Acting on the Block
The forces acting on the block are:
1. The gravitational force \( Mg \) acting downwards.
2. The tension \( T \) in the string acting upwards.

### Step 3: Apply Newton's Second Law
Since the block is accelerating, 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:
\[ T - Mg = Ma \]
Where \( a \) is the acceleration of the block.

### Step 4: Relate Work Done to Change in Kinetic Energy
According to the work-energy theorem, the work done on the block is equal to the change in kinetic energy. The work done by the tension \( T \) and the gravitational force \( Mg \) can be expressed as:
\[ \text{Work Done} = T \cdot d - Mg \cdot d \]
Where \( d \) is the distance moved by the block in the direction of the forces.

### Step 5: Calculate the Work Done
We know that the change in kinetic energy is 20 J:
\[ T \cdot d - Mg \cdot d = 20 \]
This can be rearranged to find the relationship between tension, gravitational force, and the distance moved.

### Step 6: Solve for Tension
From the previous equation, we can express the tension in terms of the gravitational force and the work done:
\[ T \cdot d = Mg \cdot d + 20 \]
Dividing through by \( d \) (assuming \( d \neq 0 \)):
\[ T = Mg + \frac{20}{d} \]

### Step 7: Analyze the Given Information
Since the force \( F \) is constant and equal to the tension \( T \) when the system is in equilibrium, we can conclude that:
\[ F = T \]
Thus, we can substitute \( T \) from the previous step:
\[ F = Mg + \frac{20}{d} \]

### Conclusion
The tension in the string is equal to the force \( F \) being applied, and the work done on the block results in an increase in kinetic energy of 20 J.