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

Answer:

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

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:

$\mu_k = \dfrac{F_k}{mg}$

$\mu_k = \dfrac{40N}{(60kg)(10m/s^2)}$

$\boxed{\mu_k = 0.07}$

Hence, the coefficient of kinetic friction is 0.07.