Question

A force acts on a body for of mass 10kg for 5sec. Then the force stops and the body describes 50m in next 8sec. The magnitude of force is?​

Answer 1

Explanation:

As we know, Applying force on a body produces acceleration. Since, uniform force is applied on the body for 5 seconds, we can assume that body moves under uniform acceleration for 5 seconds.

Let acceleration be = a m/(s)^2

Time = 5 s

Initial velocity = 0 m/s

Final velocity = v m/s

The body then moves with velocity v acquired due to motion due to force, for 8 s covering a distance of 50 m.

Therefore,

v = 50/8 m/s as

Speed = Distance/Time

Now for motion due to force, we know that

v = u + at

where v = final velocity

u = initial velocity

a = acceleration

t = time

Putting the values we get

50/8 = 0 + a(5)

50/8 = 5a

a = 10/8 m/(s)^2

Now,

Force = Mass × Acceleration

Force = 10 × 10/8 = 100/8 N ~ 12.5 N