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A museum curator moves artifacts into place on various different display surfaces. If the curator moves a 145 kg aluminum sculpture across a horizontal steel platform with a force of 668N, what is the coefficient of kinetic friction?

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meerkat18

we are given the mass of an aluminum sculpture which is 145 kg and a horizontal force equal to 668 Newtons. The coefficient of friction can be determined by dividing the horizontal force by the weight of the object. In this case, 668 N / 145 * 9.8 equal to coeff of friction of 0.47

Answer:

\(\mu_k = 0.47\)

Explanation:

As we know that the mass of the curator is

M = 145 kg

now the Force of kinetic friction on the curator is given as

\(F_k = \mu_k F_n\)

now we know that the normal force on the curator is counter balanced by the weight of the curator

so we can say

\(F_n = mg\)

\(F_n = (145)(9.8)\)

\(F_n = 1421 N\)

Now from above formula we have

\(668 = \mu_k (1421)\)

\(\mu_k = \frac{668}{1421}\)

\(\mu_k = 0.47\)

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