Question

If an object is moving at `120 m/s` over a surface with a kinetic friction coefficient of `u_k=12 /g`, how much time will it take for the object to stop moving?

Answer 1

The object will stop in 10 seconds. Complete solution follows...

Explanation:

We must start by determining the acceleration of the body, which we do by considering the force picture. Since the only horizontal force is friction, Newton's second law becomes

`F_"net" = F_f=ma`

where `F_f = muF_N = mu mg`

(if the motion is on a horizontal surface, it will be true that `F_N = mg`)

Combining these two equations, we get

`ma=mumg` or simply `a=mug`

and, since we are told that `mu=12/g`, we can express the acceleration as

`a=12m/s^2` (actually `-12 m/s^2` as teh object will be slowing down)

Now use the equation of motion `v_f = v_i +aDeltat`

Since `v_f=0, v_i=120 and a = -12,` we get

`0=120-12Deltat`

so that `Deltat` must be 10 seconds.