Answer to Question #166539 in Mechanics | Relativity for buddy

Question

At a single-vehicle accident scene, the police measured the length of a 760 kg truck’s skid marks to be 11 m up to a telephone pole, which was struck. They know that the kinetic coefficient of friction for dry pavement is 0.63. How fast was the truck moving when it put on its brakes if it hit the telephone pole going 0.8 m/s?

Answer 1

We can find the initial veloctiy of the truck from the work-kinetic energy theorem:

$$\Delta KE=W,$$ $$KE_f-KE_i=-F_{fr}d,$$ $$\dfrac{1}{2}mv_f^2-\dfrac{1}{2}mv_i^2=-\mu_kmgd,$$ $$v_i^2=2\mu_kgd+v_f^2,$$ $$v_i=\sqrt{2\mu_kgd+v_f^2},$$ $$v_i=\sqrt{2\cdot0.63\cdot9.8\ \dfrac{m}{s^2}\cdot11\ m+(0.8\ \dfrac{m}{s})^2}=11.7\ \dfrac{m}{s}.$$