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 610.0 J of potential energy with respect to street level.

Answer 1

KE = 182.24 J of energy

KE = 610 J of energy

Step-by-step explanation:

So, what the person above did is correct however they found the velocity as the brick falls from a building. The question I have on my homework is what is the kinetic energy as it is falling from 35 ft above the ground. The answer is to take the velocity that they found above which is 30.97 and put it into this formula.

KE = 1/2 * m * v^2 or (1/2 * .38 * 30.97^2)

KE = 182.24 J of energy

The second part of the question is finding the KE the moment before the brick hits the ground.

The short answer is its 610 J

Remember that at this moment, PE is equal to 0 as it is all converted to KE.

Answer 2

The correct answer is 30,97 ft/s.

Step-by-step explanation:

I looked for the problem and found the situation in which to calculate the speed of the brick.

"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?"

Initially, the kinetic energy is zero and the potential energy is equal to 610 J. As the energy is conserved, we must calculate the speed of the brick when it is 35 ft above street level.

First we calculate the mass of the brick at the starting point.

E = m * g * h

610 J = m * 32.16
`(ft)/(s^(2) )` * 50 ft

m =
`(610 J)/(1608 (ft^(2) )/(s^(2) ) )`

m = 0,38

Now we must calculate the speed when the brick is 35 ft above street level.

E = 610 J

E =
`(1)/(2)` * m *
`v^(2)` + m * g * h

610 J =
`(1)/(2)` * 0.38 *
`v^(2)` + 0.38 * 32.16
`(ft)/(s^(2) )` * 35 ft

`v^(2)` =
`((610 J - 427,73 J ))/(((1)/(2) * 0,38))`

v =
`\sqrt{(959,32ft^(2) )/(s^(2) ) }`

v = 30.97 ft/s

The speed of the brick when it falls and is 35 ft above street level is 30.97 ft/s.

Have a nice day!