230k views
1 vote
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?

User HiQ CJ
by
8.6k points

2 Answers

2 votes

Answer:


\mu_k = 0.47

Step-by-step 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 = (668)/(1421)


\mu_k = 0.47

User Zarigani
by
7.6k points
3 votes
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
User Joe Borg
by
8.1k points