At time `t=0s` particle starts moving along the `x-` axis. If its kinetic energy increases uniformly with time `'t'`, the net force acting on it must be proportional to

Options

A. (a)constant
B. (b)proportional to t
C. (c)inversely proportional to `t^(2)`
D. (d)proportional to` 1//sqrt(t)`

Explanation

To solve the problem, we need to analyze the relationship between kinetic energy, force, and acceleration. Let's break it down step by step.

### Step-by-Step Solution:

1. **Understanding Kinetic Energy**:
The kinetic energy (KE) of a particle is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the particle and \( v \) is its velocity.

2. **Given Condition**:
We are told that the kinetic energy increases uniformly with time \( t \). This means we can express kinetic energy as:
\[ KE = bt \]
where \( b \) is a constant.

3. **Relating Kinetic Energy to Velocity**:
From the kinetic energy formula, we can set:
\[ \frac{1}{2} mv^2 = bt \]
Rearranging this gives:
\[ v^2 = \frac{2bt}{m} \]
Taking the square root of both sides, we find:
\[ v = \sqrt{\frac{2bt}{m}} \]

4. **Finding Acceleration**:
Acceleration \( a \) is defined as the rate of change of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
To find \( a \), we differentiate \( v \):
\[ a = \frac{d}{dt}\left(\sqrt{\frac{2bt}{m}}\right) \]
Using the chain rule, we get:
\[ a = \frac{1}{2} \left(\frac{2b}{m}\right)^{1/2} t^{-1/2} = \frac{\sqrt{2b}}{2\sqrt{m}} t^{-1/2} \]

5. **Using Newton's Second Law**:
According to Newton's second law, the net force \( F \) acting on the particle is given by:
\[ F = ma \]
Substituting for \( a \):
\[ F = m \cdot \frac{\sqrt{2b}}{2\sqrt{m}} t^{-1/2} \]
Simplifying this gives:
\[ F = \frac{\sqrt{2bm}}{2} t^{-1/2} \]

6. **Conclusion**:
From the final expression for force, we see that:
\[ F \propto t^{-1/2} \]
This indicates that the net force acting on the particle is inversely proportional to the square root of time.

### Final Answer:
The net force acting on the particle must be proportional to \( t^{-1/2} \).