Question

you apply constant horizontal force of 80N to a wooden crate on a smooth horizontal floor.The crate start from rest and moves 11m in 5 seconds.If you stop pushing at the end of 5 seconds,how far does the crate move in the next 5 seconds?

Answer 1

Answer

By second equation of motion

$s=ut+\frac{1}{2}at^2\\11=0(5) +\frac{1}{2}a(5) ^2\\a=0.88m/s^2$

Velocity after first 5second

$v=0.88(5) =4.4m/s$

When force will removed then acceleration will be zero

So distance travelled in next 5 second

$s=4.4\times5=22m$