Answer
(a) $-587J$
(b) $0.284$
Work Step by Step
(a) We can find the required work done as follows:
$W=K.E_f-K.E_i$
$\implies W=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$
As the player comes to rest, $v_f=0$
$\implies W=0-\frac{1}{2}mv_i^2$
We plug in the known values to obtain:
$W=-\frac{1}{2}(62.0)(4.35)^2$
$W=-587J$
(b) The coefficient of kinetic friction can be determined as
$W=-(\mu_kmg)d$
This can be rearranged as:
$\mu_k=\frac{-W}{mgd}$
We plug in the known values to obtain:
$\mu_k=\frac{-587}{(62.0)(9.81)(3.40)}=0.284$
