Question

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?

Answer 1

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 2

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

$668 = \mu_k (1421)$

$\mu_k = \frac{668}{1421}$

$\mu_k = 0.47$