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 Set1 Q11
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 eqn, 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)
The velocity acquired by the body 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