Question

How much work would it take to horizontally accelerate an object with a mass of `6 kg` to `7 m/s` on a surface with a kinetic friction coefficient of `3`?

Answer 1

The work is `=147J`

Explanation:

The coefficient of kinetic friction is

`mu_k=F_r/(N)`

The normal force is

`(N)=mg=6g N`

The frictional force is

`F_r=mu_k*(N)=3*6g=18gN`

According to Newton's Second Law, the acceleration is

`F=ma`

`a=F/m=18g/6=3g`

To find the distance, we apply the equation of motion

`v^2=u^2+2as`

`u=0`

`s=v^2/(2a)=7^2/(2*3g)m`

The work is

`W=Fs=18g*49/(6g)=147J`