Question

What is the kinetic energy of a 15 kg object travelling at 6 m/s. What is the height of the height where the Object must be dropped if it will be raised up ?

Answer 1

(a)

$$KE=\dfrac{1}{2}mv^2,$$ $$KE=\dfrac{1}{2}\cdot15\ kg\cdot(6\ \dfrac{m}{s})^2=270\ J.$$

(b) We can find the height from the law of conservation of energy:

$$PE=KE,$$ $$mgh=KE,$$ $$h=\dfrac{KE}{mg}=\dfrac{270\ J}{15\ kg\cdot9.8\ \dfrac{m}{s^2}}=1.84\ m.$$