A force acts on the body of mass 50 kg, for 10 second. When the force stops acting on the body, the body covers 80 m in the next 10 second. What is the magnitude of the force?

Options

A. 40N
B. 50N
C. 30N
D. 60N

Explanation

To solve the problem, we need to find the magnitude of the force acting on a body of mass 50 kg, which acts for 10 seconds, after which the body covers 80 m in the next 10 seconds when the force stops acting.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Mass of the body, \( m = 50 \, \text{kg} \)
- Time during which the force acts, \( t_1 = 10 \, \text{s} \)
- Distance covered after the force stops acting, \( d = 80 \, \text{m} \)
- Time during which the body covers this distance, \( t_2 = 10 \, \text{s} \)

2. **Determine the Velocity After the Force Stops Acting:**
- Since the body covers 80 m in 10 seconds after the force stops acting, we can calculate the velocity during this interval.
\[ v = \frac{d}{t_2} = \frac{80 \, \text{m}}{10 \, \text{s}} = 8 \, \text{m/s} \]

3. **Understanding Motion Before the Force Stops Acting:**
- The body starts from rest, so the initial velocity \( u = 0 \).
- Let \( a \) be the acceleration while the force is acting.
- The final velocity \( v \) after 10 seconds can be expressed using the equation of motion:
\[ v = u + at_1 \implies 8 \, \text{m/s} = 0 + a \cdot 10 \, \text{s} \]
- Thus, we can solve for acceleration \( a \):
\[ a = \frac{8 \, \text{m/s}}{10 \, \text{s}} = 0.8 \, \text{m/s}^2 \]

4. **Calculate the Force:**
- Using Newton's second law, the force \( F \) can be calculated as:
\[ F = m \cdot a = 50 \, \text{kg} \cdot 0.8 \, \text{m/s}^2 = 40 \, \text{N} \]

5. **Conclusion:**
- The magnitude of the force acting on the body is \( 40 \, \text{N} \).