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viva [34]
4 years ago
6

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

Physics
1 answer:
dmitriy555 [2]4 years ago
6 0

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.

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