Question

How much work would it take to horizontally accelerate an object with a mass of `2 kg` to `9 ms^-1` on a surface with a kinetic friction coefficient of `1`?

Answer 1

The total work done will include `81 J` for the acceleration and `19.6d J` to overcome the frictional force, for a total of `W = (81+19.6d) J.`

Explanation:

Some of the work done will be done accelerating the object. Its kinetic energy is `0` when it is at rest and as its velocity rises to `9 ms^-1` this rises to:

`E_k = 1/2mv^2 = 1/2*2*9^2 = 81 J`

The change in kinetic energy is simply the work done in accelerating the object, but there is also work done against the frictional force.

The fact that the friction coefficient is 1 means that the frictional force acting will be the same as the normal force, which in this case is the weight force of the object:

`F = mg = 2*9.8=19.6 N`

This is the force that needs to be overcome. To calculate the work done, normally we would use the formula:

`W = Fd` where `d` is the distance traveled.

In this case we don't know the distance, so we may just need to leave it in terms of `d`:

`W = 81 + 19.6*d = (81 + 19.6d) J` where d is the distance through which the object moves while accelerating.