A particle of mass m is moving with speed u. It is stopped by a force F in distance x if the stopping force is 4 F then

Options

A. work done by stopping force in second case will be same as that in first case.
B. work done by stopping force in second case will be 2 times of that in first case.
C. work done by stopping force in second case will be `(1)/(2)` of that in first case.
D. work done by stopping force in second case will be `(1)/(4)` of that in first case.

Explanation

To solve the problem, we need to analyze the work done by the stopping forces in both cases and relate it to the change in kinetic energy of the particle.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions:**
- A particle of mass \( m \) is moving with an initial speed \( u \).
- It is stopped by a force \( F \) over a distance \( x \).

2. **Calculating Work Done in the First Case:**
- The work done \( W_1 \) by the stopping force \( F \) in the first case can be calculated using the formula:
\[ W_1 = F \cdot x \]

3. **Using the Work-Energy Principle:**
- According to the work-energy principle, the work done on the particle is equal to the change in kinetic energy:
\[ W = \Delta KE = KE_{\text{final}} - KE_{\text{initial}} \]
- The initial kinetic energy \( KE_{\text{initial}} \) is given by:
\[ KE_{\text{initial}} = \frac{1}{2} m u^2 \]
- Since the particle comes to rest, the final kinetic energy \( KE_{\text{final}} \) is 0. Thus:
\[ W_1 = 0 - \frac{1}{2} m u^2 = -\frac{1}{2} m u^2 \]

4. **Calculating Work Done in the Second Case:**
- In the second case, the stopping force is \( 4F \).
- Let \( x' \) be the distance over which the particle is stopped by the force \( 4F \).
- The work done \( W_2 \) in this case is:
\[ W_2 = 4F \cdot x' \]

5. **Applying the Work-Energy Principle Again:**
- The change in kinetic energy remains the same since the particle is stopped from the same initial speed \( u \):
\[ W_2 = -\frac{1}{2} m u^2 \]

6. **Equating the Work Done:**
- Since both cases result in the same change in kinetic energy, we equate the work done:
\[ 4F \cdot x' = -\frac{1}{2} m u^2 \]
- From the first case, we know:
\[ F \cdot x = -\frac{1}{2} m u^2 \]
- Therefore, we can relate \( x' \) to \( x \):
\[ 4F \cdot x' = F \cdot x \implies x' = \frac{x}{4} \]

7. **Conclusion:**
- The work done by the stopping force in both cases is equal, as both bring the particle to rest from the same initial speed \( u \).
- Therefore, the correct answer is that the work done by the stopping force in the second case will be the same as that in the first case.

### Final Answer:
- The work done by the stopping force in the second case will be the same as that in the first case.