Question

A brick of mass 2 kg is dropped from a rest position 5 m above the ground. what is its velocity at a height of 3 m above the ground?

Answer 1

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

Using the ground as reference point, the mechanical energy of the brick when it is at 5 m from the ground is just potential energy (because the brick is initially at rest, so it doesn't have kinetic energy):
$E= U = mgh=(2 kg)((9.81 m/s^2)(5 m)=98.1 J$

when the brick is at h'=3 m from the ground, its mechanical energy is now sum of kinetic energy and potential energy:
$E= K+U= \frac{1}{2} mv^2 + mgh'$

where v is the velocity of the brick. Since E is conserved, it must be equal to the initial energy (98.1 J), so we can solve this equation to find v:
$v= \sqrt{ \frac{2(E-mgh')}{m} }=6.3 m/s$