Question

An 18 kg curling stone sliding on a sheet of ice hits a rough patch on the ice where the coefficient of friction increases from 0.13 to 0.28. It enters this rough patch with a speed of 2.2 m/s and leaves with
a speed of 1.4 m/s. How long is the rough patch?

What are we suppose to do in this question?

Answer 1

work done = force * distance = change in kinetic energy

F = -.28 m g = -.28 * 18 * 9.81
work = F * x

final - initial Ke

= (1/2) *18 * (.13^2 - .28^2)

so
-.28 * 9.81 = (1/2) * (.13^2 - .28^2)

Answer 2

forgot x
-.28 * 9.81 * x = (1/2) * (.13^2 - .28^2)