A constant force acts on an object of mass 5 kg for a duration of 2 s. It increases the object’s velocity from 3 m s–1 to 7 m s-1. Find the magnitude of the applied force. Now, if the force was applied for a duration of 5 s, what would be the final velocity of the object?

Solution:
We have been given that u = 3 m s^–1 and v = 7 m s-1 , t = 2 s and m = 5 kg.

we have,
F = m ( v – u ) / t
Substitution of values in this relation gives
F = 5 kg ( 7 m s^-1 – 3 m s^-1) / 2 s = 10 N.
Now, if this force is applied for a duration of 5 s (t = 5 s),

then the final velocity can be calculated by rewriting as v = u + Ft / m
On substituting the values of u, F, m
and t, we get the final velocity, v = 13 m/s² .