At the top of the hill, the skier has only potential energy, because its speed is zero (he starts from rest), so his mechanical energy is
[tex]E_i = U=mgh = (60 kg)(9.81 m/s^2)(50 m)=29430 J[/tex]
During his descent, the work done by the friction is [tex]W=-6.0 kJ=-6000 J[/tex]. This means that the skier loses 6000 J of energy during his descent, so its final mechanical energy at the bottom is
[tex]E_f = E_i -6000 J=23430 J[/tex]
At the bottom of the hill, the mechanical energy is only kinetic energy K, because now the altitude of the skier is zero: h=0, so the potential energy is zero (U=0). So,
[tex]E_f = K= \frac{1}{2}mv^2 [/tex]
And so, from this we find the speed of the skier at the bottom of the hill:
[tex]v= \sqrt{ \frac{2K}{m} } = \sqrt{ \frac{2\cdot 23430 J}{60 kg} }=14.0 m/s [/tex]
