A block of mass m is kept on the ground. A vertically upward force F = kt (K is a positive constant) where 't' is in seconds , starts to act on it at t = 0

Options

A. the block looses contact with ground at t = mgK
B. For `0 lt t le (mg)/k` , normal reaction decrease linearly with time but force exerted by body on ground also decreases
C. For `0 lt t le (mg)/k` , normal reaction decrease linearly with time but force exerted by body on ground is a constant
D. At `t=(2mg)/k` , velocity of the body is zero

Answer

To solve the problem step by step, we need to analyze the forces acting on the block of mass \( m \) when a vertically upward force \( F = kt \) is applied.

### Step 1: Identify the forces acting on the block
- The block experiences two main forces:
- The downward gravitational force: \( F_g = mg \)
- The upward force: \( F = kt \)

### Step 2: Write the equation of motion
- According to Newton's second law, the net force acting on the block can be expressed as:
\[ F_{\text{net}} = F - F_g = kt - mg \]
- The normal force \( N \) also acts upwards. Therefore, we can write:
\[ N + kt = mg \]

### Step 3: Determine when the block loses contact with the ground
- The block loses contact with the ground when the normal force \( N \) becomes zero:
\[ N = 0 \implies kt = mg \]
- Rearranging gives:
\[ t = \frac{mg}{k} \]

### Step 4: Analyze the normal force over time
- As time \( t \) increases, the upward force \( kt \) increases linearly. Consequently, the normal force \( N \) must decrease to maintain the equilibrium:
\[ N = mg - kt \]
- This indicates that the normal force decreases linearly with time until it reaches zero.

### Step 5: Determine the force exerted by the block on the ground
- The force exerted by the block on the ground is equal to the normal force \( N \):
\[ F_{\text{exerted}} = N = mg - kt \]
- As \( kt \) increases, \( N \) decreases, which means the force exerted by the block on the ground also decreases.

### Step 6: Analyze the velocity of the block at a specific time
- To find the velocity of the block at \( t = \frac{2mg}{k} \):
- The acceleration \( a \) at any time \( t \) can be expressed as:
\[ a = \frac{F_{\text{net}}}{m} = \frac{kt - mg}{m} = \frac{kt}{m} - g \]
- Substituting \( t = \frac{2mg}{k} \):
\[ a = \frac{k \cdot \frac{2mg}{k}}{m} - g = 2g - g = g \]
- Since the acceleration is \( g \), the block is not at rest, and its velocity cannot be zero.

### Conclusion
- The block loses contact with the ground at \( t = \frac{mg}{k} \).
- The normal force decreases linearly with time.
- The force exerted by the block on the ground also decreases.
- The velocity of the block at \( t = \frac{2mg}{k} \) is not zero.

### Summary of Correct Options
1. The block loses contact with the ground at \( t = \frac{mg}{k} \) (Correct).
2. The normal reaction decreases linearly with time (Correct).
3. The force exerted by the body on the ground decreases (Correct).
4. The velocity of the body at \( t = \frac{2mg}{k} \) is not zero (Incorrect).