Answer to Question #216764 in Physics for Sadia Shafique

Question

1) A 5 kg watermelon is placed on the edge of a balcony 43.0 m high.

a) What is the watermelon's gravitational potential energy?

b) If all of the gravitational potential energy is converted into kinetic energy after the box is pushed off the balcony, how fast will the watermelon be moving when it hits the ground?

Answer 1

(a)

$$GPE=mgh=5\ kg\cdot9.8\ \dfrac{m}{s^2}\cdot43\ m=2107\ J.$$

(b)

$$GPE=KE=\dfrac{1}{2}mv_f^2,$$ $$v_f=\sqrt{\dfrac{2GPE}{m}}=\sqrt{\dfrac{2\cdot2107\ J}{5\ kg}}=29\ \dfrac{m}{s}.$$