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. A

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)

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Nick 4 years 2021-07-31T03:13:33+00:00 1 Answers 89 views 0

Answers ( )

  1. Bo
    0
    2021-07-31T03:15:32+00:00
    Reply

    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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