A force acts for 10 s on a stationary body of mass 100 kg after which the force ceases to act. The body moves through a distance of 100 m in the next 5 s.

Calculate (i) the velocity acquired by the body,

(ii) the acceleration produced by the force,

and (iii) the magnitude of the force.

## Solution of class 9 Set2 Q1

Solution

Mass of the object= 100 kg

As the force acts on the body for 10 secs, that means the body accelerates for 10 sec.

Time duration =t= 10 s

As per the problem statement, the initial velocity u =0

Let the velocity acquired by the body due to acceleration in 10 secs= final velocity = v

Let acceleration =a

So as per equation, v= u + at

=> v = 0 + a.10

=> v = 10 a ……………….(a)

Now, when the force is removed, the object continues to move with uniform velocity v.

And as per the question, it covers 100 m in 5 sec with this velocity v.

As here with uniform velocity, Distance(s) = v.t

so, 100 = v. 5

=> v = 100/5 = 20 m/s…………….(b)

So the value of the velocity acquired is 20 m/s

Now merging (a) and (b) we get,

v= 10 a

=> 20 = 10 a

so, a = 20/10 = 2 m/s^2

The value of acceleration produced by the force=2 m/s^2

The force = F = mass . acceleration = 100.2 N = 200 N