Question

A force of `5N` changes the velocity of a body from `10 ms^(-1)` to `20 ms^(-1)` in `5 sec`. How much force is required to bring about the same change in `2 sec`?

Answer 1

From `F_(1) = (dp)/(dt_(1))and F_(2) = (dp)/(dt_(2)) `
` (F_(2))/(F_(1))= dt_(1)/(dt_(2)), F_(2) = F_(2) = F_(1) xx dt_(1)/(dt_(2)) = (5xx 5 )/(2) = 12 .5 N ` .