A 6.0-kg block slides from point A down a frictionless curve to point B. After the block passes point B, a friction force opposes the motion of the block so that it comes to a stop 2.5 m from B. Calculate the coefficient of kinetic friction between the block and the surface after position B.

Question

Quizplus · Accepted Answer

The initial energy of the block is transformed into work done by the friction force after point B, therefore:
(1/2)mv^2 = F_k x
where m = 6.0 kg, v = sqrt(2gh) where h = distance from A to B = 2.5 m, x = distance from point B to stop = 2.5 m, and F_k is the kinetic friction force. 
Solving for F_k and substituting the expression for F_k from the kinetic friction equation: F_k = mu_k * N, where N is the normal force, we obtain:
(1/2)m(v^2) = (mu_k * N) * x
N = mg, where g is the acceleration due to gravity
Substituting and simplifying yields:
mu_k = (1/2)(v^2)/(gx) = (1/2)((sqrt(2gh))^2)(gx)) = 0.40, which is less than or equal to the coefficient of static friction. Therefore, the answer is B.

A 60-Kg Block Slides from Point a Down a Frictionless Curve