A constant force acts on an object of mass 5kg for a duration of 2s.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 5s,What would be the final velocity of the object ?

Explanation

Here,u=3 `ms^(-1)` and v=7 m `s^(-1)`
t=2s and m=5kg
`F=(m(v-u))/(t)`
Substitution of values in this relation gives
`F=(5kg(7 ms^(-1)-3 ms^(-1)))/(2s)`
`therefore F=10N`
Now, if this force is applied for a duration of 58 t = 5s), then the final velocity can be calculated by rewriting,
`v=u+(ft)/(m)`
On substituting the values of u. F. m and t, we get the final velocity v = 13 m`s^(-1)`