Question

A brick lies perilously close to the edge of the flat roof of a building. The roof edge is 50 ft above street level, and the brick has 710.0 J of potential energy with respect to street level. Someone edges the brick off the roof, and it begins to fall. What is the brick’s kinetic energy when it is 35 ft above street level? What is the brick’s kinetic energy the instant before it hits the street surface?

Answer 1

The kinetic energy just before it hits the street is 710.0 J. Please check the explanation below to help you remember the steps.

Step-by-step explanation:

The total mechanical energy of the brick (potential energy + kinetic energy) is conserved as it falls, assuming no air resistance. This means that the sum of the potential energy and kinetic energy at any point during the fall is equal to the total mechanical energy at the start, which is 710.0 J.

The potential energy (PE) of the brick at a height h is given by the formula PE = mgh, where m is the mass of the brick, g is the acceleration due to gravity (9.8 m/s² or 32.2 ft/s²), and h is the height above the ground.

1. When the brick is 35 ft above the street level, its potential energy is:

PE = m * 32.2 ft/s² * 35 ft

We can set this equal to 710.0 J - KE, where KE is the kinetic energy at this height, and solve for KE:

KE = 710.0 J - PE

2. Just before the brick hits the street, its potential energy is zero (because its height is zero), so all of its mechanical energy is kinetic energy. Therefore, the kinetic energy just before it hits the street is 710.0 J.