Question

A 60kg block rests on rough horizontal ground. A rope is

attached to the block and is pulled with a force of 220N to the
left. As a result, the block accelerates at 3 m/s2. The coefficient
of kinetic friction Hk between the block and the ground is
(round to the nearest hundredth)

Answer 1

The coefficient of kinetic friction is
`\mu_k = 0.07.`

Step-by-step explanation:

Let us call
`F_k` the force of friction, then we know that
`220N - F_k` is what has caused the acceleration
`a =3m/s^2`:

`220-F_k =ma`

`220-F_k =(60kg)(3m/s^2)`

`220-F_k =180`

`F_K = 40N`

Now this frictional force relates to the coefficient of kinetic friction
`\mu_k` by

`F_k = \mu_k N`

where
`N=mg` is the normal force.

Putting in numbers and solving for
`\mu_k` we get:

`\mu_k = (F_k)/(mg)`

`\mu_k = (40N)/((60kg)(10m/s^2))`

`\boxed{\mu_k = 0.07}`

Hence, the coefficient of kinetic friction is 0.07.