A force acts for 20 s on a body of mass 20 kg, starting form rest, after which the force ceases and then body describes 50 m in the next 10 s. The value of force will be:

Options

A. `40 N`
B. `20 N`
C. ` 5 N`
D. `10 N`

Answer

To solve the problem step by step, we can follow these steps:

### Step 1: Understand the problem
We have a body of mass \( m = 20 \, \text{kg} \) that starts from rest and is acted upon by a force for \( t_1 = 20 \, \text{s} \). After the force ceases, the body moves for an additional \( t_2 = 10 \, \text{s} \) and covers a distance of \( d = 50 \, \text{m} \).

### Step 2: Calculate the final velocity after the force acts
Since the body starts from rest, the initial velocity \( u = 0 \). We need to find the acceleration \( a \) during the time the force is acting.

Using the kinematic equation:
\[ d = ut + \frac{1}{2} a t^2 \]
For the first 20 seconds, the distance covered is not given, but we can express it in terms of acceleration:
\[ d_1 = 0 + \frac{1}{2} a (20)^2 = 200a \]

### Step 3: Determine the velocity at the end of 20 seconds
The final velocity \( v \) at the end of 20 seconds can be calculated using:
\[ v = u + at = 0 + a \cdot 20 = 20a \]

### Step 4: Analyze the motion after the force ceases
After the force ceases, the body moves with a constant velocity \( v \) for \( t_2 = 10 \, \text{s} \). The distance covered in this time is given by:
\[ d = vt_2 = (20a)(10) = 200a \]
We know from the problem that this distance is 50 m:
\[ 200a = 50 \]
Solving for \( a \):
\[ a = \frac{50}{200} = \frac{1}{4} \, \text{m/s}^2 \]

### Step 5: Calculate the force
Now that we have the acceleration, we can find the force using Newton's second law:
\[ F = ma = 20 \cdot \frac{1}{4} = 5 \, \text{N} \]

### Final Answer
The value of the force is \( \boxed{5 \, \text{N}} \).

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