Question

A 1300-kg car going 40 m/s applies its breaks and skids to rest. If the friction force between the sliding tires and the pavement is 7,000 N. How far does the cars skid before coming to rest?

Answer 1

$$\Delta KE=W,$$ $$\dfrac{1}{2}mv_f^2-\dfrac{1}{2}mv_i^2=-F_{fr}d,$$ $$d=\dfrac{-\dfrac{1}{2}mv_i^2}{-F_{fr}},$$ $$d=\dfrac{-\dfrac{1}{2}\cdot1300\ kg\cdot(40\ \dfrac{m}{s})^2}{-7000\ N}=148.6\ m.$$