Problem 6.48

6.48. a)

The energy lost after one revolution is given by the change in kinetic energy.

$W = \Delta K = \frac{m(v_{f}^{2} - v_{i}^{2})}{2}$

W = -5.60 J which is lost to friction.

b)

If we want the coefficient of kinetic friction, let's figure in exactly what the work done by friction is.

$W = F_{f} d ~=~ -\mu_{k} mg(2\pi r)$

therefore, $\mu_{k}$ is given by

$\mu_{k} = \frac{5.6}{(0.4)(9.8)(2\pi)(1.5)} = .152$

c)

Now, we want the work done after N revolutions to be equal to the initial kinetic energy. To do this we find

$K_{i} = \frac{1}{2} (0.4)(8)^{2} ~=~ 12.8 J$

So equating the work done by friction in N revolutions to this value we find

$W ~=~ \mu_{k}mg(2\pi r N)$

$12.8 J ~=~ (.152)(.4)(9.8)(2\pi) N$

or N = 2.28 revolutions.