A horizontal 50 N force acts on a 2 kg crate which is at rest on a smooth horizontal surface. At the instant the particle has gone 2 m, the rate at which the force is doing work is :

Options

A. 2.5 W
B. 25 W
C. 100 W
D. 500 W

Explanation

To solve the problem, we need to find the rate at which the force is doing work on the crate after it has moved 2 meters. This rate is defined as power.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Force (F) = 50 N
- Mass (m) = 2 kg
- Displacement (s) = 2 m

2. **Calculate the Acceleration:**
- According to Newton's second law, acceleration (a) can be calculated using the formula:
\[ a = \frac{F}{m} \]
- Substituting the values:
\[ a = \frac{50 \, \text{N}}{2 \, \text{kg}} = 25 \, \text{m/s}^2 \]

3. **Determine the Final Velocity:**
- We can use the kinematic equation to find the final velocity (v) after the crate has moved 2 meters. The equation is:
\[ v^2 = u^2 + 2as \]
- Since the crate starts from rest, the initial velocity (u) = 0. Thus:
\[ v^2 = 0 + 2 \cdot 25 \, \text{m/s}^2 \cdot 2 \, \text{m} \]
\[ v^2 = 100 \]
\[ v = \sqrt{100} = 10 \, \text{m/s} \]

4. **Calculate the Instantaneous Power:**
- The instantaneous power (P) can be calculated using the formula:
\[ P = F \cdot v \]
- Substituting the values we have:
\[ P = 50 \, \text{N} \cdot 10 \, \text{m/s} = 500 \, \text{W} \]

### Final Answer:
The rate at which the force is doing work is **500 Watts**.

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