Question

A force of 300N[S] is applied to a 50kg box on a rough floor. The frictional force is 200N. The acceleration is 2m/s^2.

if the 300N force is removed after 4.0s, how long will it take the box to come to rest?

please show ALL your work!

Answer 1

We want to know how far this box will move after the force stops. This is called the displacement and it can be calculated using:


`Displacement = Final\ Velocity * Time`

We do not know the final velocity of the box however this can be calculated using:


`Final\ Velocity = Initital\ Velocity + (Acceleration * Time)`

When we put our values into this equation we get:


`Final\ Velocity = 0 + (2m/s^2 * 4.0s)`

`Final\ Velocity = 8m/s`

We can put this value into the Displacement equation to get


`Displacement = 8m/s * 4s`

`Displacement = 32m`

This means that the total difference in space between where the box stopped receiving a force and came to halt is 32m. Presuming the floor was completely flat, then the only way the box could move is forwards. This means that there is a 32m movement forwards between the two places which means the box took 32m to come to rest.