Question

a man who with 55kg, on a skateboard. He had an initial speed of 2.77m/s when descending the stairs at a height of 2m. Calculate the man’s potential energy when his speed is 6m/s. 

Answer 1

Let's write the law of conservation of mechanical energy:

$$KE_i+PE_i=KE_f+PE_f,$$ $$\dfrac{1}{2}mv_1^2+mgh_1=\dfrac{1}{2}mv_2^2+PE_f,$$ $$PE_f=\dfrac{1}{2}m(v_1^2-v_2^2)+mgh_1,$$ $$PE_f=\dfrac{1}{2}\cdot 55\ kg\cdot((2.77\ \dfrac{m}{s})^2-(6\ \dfrac{m}{s})^2)+55\ kg\cdot 9.8\ \dfrac{m}{s^2}\cdot 2\ m=299\ J.$$