A person drops a brick from the top of a building. The height of the building is 400 m and the mass of the brick is 2.00 kg. What will be the speed of the brick right before it touches the ground? Use g = 10.0 m/s².

Question

Quizplus · Accepted Answer

The potential energy of the brick at the top of the building is given by:

PE = mgh = (2.00 kg)(10.0 m/s^2)(400 m) = 8000 J

At the bottom of the building, all of this potential energy will be converted into kinetic energy:

KE = 1/2 mv^2

Rearranging this equation gives:

v = sqrt(2KE/m)

Substituting in the values for KE and m and solving for v gives:

v = sqrt(2(8000 J)/(2.00 kg)) = 89.4 m/s

Therefore, the speed of the brick right before it touches the ground is approximately 89.4 m/s.

A Person Drops a Brick from the Top of a Building