## Physics: Principles with Applications (7th Edition)

The sled’s mechanical energy is conserved. Take y = 0 at the bottom. The initial KE + initial PE equals the final KE + final PE. The sled comes to a momentary stop at the end. $$\frac{1}{2}mv_i^2 + mgy_i=\frac{1}{2}mv_f^2 + mgy_f$$ $$\frac{1}{2}mv_i^2 + 0 = 0 + mgy_f$$ $$v_i=\sqrt{2gy_f}=4.89 m/s$$