Work due to friction with an ice skater?

Question

Any ideas opn how to do this?? please try to answer both questions.

A skater with a mass of 52.0 kg moving at 2.5m/s glides to a stop over a distance of 24meters. How much work did the friction of the ice do to bring the skater to a stop? How much work would the skater have to do to speed up to 2.5m/s again.

Answer 1

The work done is the same as the kinetic energy lost by the skater, i.e. W = E = mv^2 / 2 = 52.0 (2.5)^2 / 2 = 1.6 * 10^2 J. The skater would have to do the same amount of work to get back to 2.5 m/s again.