Question

How much work would it take to horizontally accelerate an object with a mass of `3 kg` to `7 m/s` on a surface with a kinetic friction coefficient of `3`?

Answer 1

`mumgDeltax+1/2mv^2`

Explanation:

Use work-energy theorem.

`W_"net" = DeltaK`

If you ask how much 'net' work is done, then the answer is

`W_"net" = 1/2mv^2 - 1/2mv_0^2 =1/2mv^2 - 0= 1/2mv^2`

However, if you ask how much work is needed to overcome the friction to achieve `v=7m/s`, then you need to consider the net force and the distance traveled (`Deltax`):

`W_"net"= F_"net" Deltax = (F_"applied force" - F_mu)Deltax =DeltaK`

`F_mu = muN= muMg` is the frictional force

Therefore,

`F_"applied force"Deltax = F_muDeltax+ DeltaK= mumg Deltax + 1/2mv^2`
The workdone by the applied force therefore does two things: to overcome the frictional work and to change the kinetic energy of the object.

However, neither x nor time t to reach `v=7m/s` are given. You need either one to solve this problem unless you meant net work done.