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

Here, `m=5kg, t=2s, u=3ms^(-1), v=7ms^(-1), F=?`
`F=ma=(m(u-v))/t = (5(7-1))/2=10N`
In the second case, `F = 10N, t = 5s, v = ?, u = 3m//s`
`a = F/m = 10/(5') = 2m//s^(2)`
From `v=u+at, v=3+2xx5=13m//s`