Question

A 50.0kg skier travels 0.200km down a hill that has a 20 degree slope. When the skier reaches the bottom of the hill, her speed is 30.0m/s. How much work is done by friction as the skier comes down the hill?

Answer 1

Work is modeled as \[\large\ force * distance * \cos \theta\]However, work done by a net force of an object/system is equal to the object/system's change in kinetic energy. Given the final speed of 30.0 m/s, apply the following equation:\[\large\ W _{net}=\Delta KE\]