A constant force acts on object of mass 5 kg for a duration of 2s It increases the object's velocity from `3m//s` to `7m//s` Find the magnitude of the applied force. Now if the force were applied for a duration of `5s` what would be the final velocity of object?

Explanation

We have been given that u = 3 m `s^(–1)`
and v = 7 m `s^(-1)`, t = 2 s and m = 5 kg.
From Eq. (9.5) 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 Eq. (9.5) as `v=u+(Ft)/(m)` On substituting the values of u, F, m and t, we get the final velocity,