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 340.0 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?

Question

Chemistry

Shaylee Preston

8 October, 17:51

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 340.0 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?

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Kendal · Answer 1

First let us lay out the formula for Potential and Kinetic Energy:

PE = mgh

KE = (mv^2) / 2

where: m=mass, v=velocity, g=gravitational acceleration, h=height

Calculating for mass using the known values of PE:

340J = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

The law of conservation of energy states that energy is neither created nor destroyed. Therefore, the change in Kinetic Energy from 50 ft to 35 ft would be equal to the change in Potential Energy from 50 ft to 35 ft. However it would be opposite in signs since one is losing while other is gaining.

KE (35ft) - KE (50ft) = - [ PE (35ft) - PE (50ft]

KE (50ft) = 0 since the brick is initially at rest

KE (35ft) = 340J - 2.27kg (9.81 m/s^2) (35ft) (1m/3.28 ft)

KE (35ft) = 102.38 J