A constant force acts on a body of mass 1 kg for 5 s and then ceases to work . In the next 5 s the body describes 100 m . Calculate the force that acted on the body.

Answer

To solve the problem step by step, we will follow the principles of kinematics and Newton's second law of motion.

### Step 1: Understand the problem
We have a body of mass \( m = 1 \, \text{kg} \) that is acted upon by a constant force for \( t = 5 \, \text{s} \). After the force ceases, the body moves for another \( 5 \, \text{s} \) and covers a distance of \( s = 100 \, \text{m} \).

### Step 2: Calculate the final velocity after the force acts
The body moves for \( 5 \, \text{s} \) after the force has stopped. We need to find the velocity \( v \) of the body at the end of the first \( 5 \, \text{s} \).

Using the formula for average velocity when distance and time are known:
\[ v = \frac{s}{t} = \frac{100 \, \text{m}}{5 \, \text{s}} = 20 \, \text{m/s} \]

### Step 3: Relate the final velocity to acceleration
We can use the kinematic equation to relate the final velocity \( v \), initial velocity \( u \), acceleration \( a \), and time \( t \):
\[ v = u + at \]
Here, the initial velocity \( u = 0 \) (the body starts from rest), and \( t = 5 \, \text{s} \). Thus, we have:
\[ 20 \, \text{m/s} = 0 + a \cdot 5 \, \text{s} \]
From this, we can solve for acceleration \( a \):
\[ a = \frac{20 \, \text{m/s}}{5 \, \text{s}} = 4 \, \text{m/s}^2 \]

### Step 4: Apply Newton's second law
According to Newton's second law, the force \( F \) acting on an object is given by:
\[ F = ma \]
Substituting the values we have:
\[ F = 1 \, \text{kg} \cdot 4 \, \text{m/s}^2 = 4 \, \text{N} \]

### Conclusion
The force that acted on the body is \( 4 \, \text{N} \).

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