Question

If a rock weighing 2,200 N is dropped from a height of 15m, what is the kinetic energy just before it hits the ground?

Answer 1

We can solve the problem by using the law of conservation of energy.

Initially, the rock has only gravitational potential energy, which is given by
$E_i = U = mgh$
where mg is the weigth of the rock (2200 N), while h is the height at which the rock has been released (h=15 m). If we calculate it, we get
$E_i = mgh=(2200 N)(15 m)=33000 J =33 kJ$

Just before hitting the ground, the rock height is zero, so its potential energy is now zero. So the total mechanical energy of the rock now is just kinetic energy:
$E_f = K_f$
however, the mechanical energy of the rock must be conserved, so
$E_i = E_f$

and so we have that the kinetic energy of the rock just before hitting the ground is equal to its initial potential energy:
$K_f = E_i =U_i = 33 kJ = 33000 J$