#### Answer

a) $W_{f_k}=-268.4J$
b) $f_k=141.3N$

#### Work Step by Step

a) As the hill makes an angle of $25^o$ with the horizontal, a distance of $1.9m$ up the hill corresponds to a $1.9\times\sin25=0.8m$ rise in height.
We assume kinetic friction is the only non-conservative force here. According to the work-energy theorem, $$W_{nc}=E_f-E_0$$ $$W_{f_k}=\frac{1}{2}m(v_f^2-v_0^2)+mg(h_f-h_0)$$
We have $m=63kg, v_0=6.6m/s, v_f=4.4m/s, h_f-h_0=0.8m, g= 9.8m/s^2$
The work done by $f_k$ on the skis is $$W_{f_k}=-268.4J$$
b) The magnitude of the kinetic frictional force can be found by $$f_k=\frac{W_{f_k}}{s\cos\theta}$$
The distance of the skis is $s=1.9m$. $f_k$ opposes the motion along its way, so $\cos\theta=\cos180=-1$
$$f_k=\frac{-268.4J}{-1.9m}=141.3N$$