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 ?

Explanation

We have been given that u = 3 `ms^(-1)` and v = 7 `ms^(-1), t = 2` s and m = 5 kg.
From Eq. we have,
`F = (m(v-u))/(t)`
Substitution of values in this relation gives
`F = (5kg(7ms^(-1) - 3ms^(-1))/(2s) = 10N`
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. as
`v = u + (Ft)/(m)`
On substituting the values of u, F, m and t, we get the final veloctiy,
`v = 13ms^(-1)`