How to Calculate Mechanical Energy Lost During a Fall?

[zelda1850]

Homework Statement

a 6.0kg concrete block is dropped from the top of a tall building. the block has fallen a distance of 55 meters and has a speed of 30 meters per srcond when it hits the ground

1) at the instant the block was released what was the gravitational potenetial energy with respect to the ground?

2)calculate the kinetic energy of the block at the point of impact?

3) how much mechanical energy was lost by the block as it fell?

4) explain what happened to the mechanical energy that was lost by the block?

Homework Equations

1) pe = mgh so 6kg x 9.81m/s2 x 55m = 323 j

2) ke = 1/2mv2 so 6kg x 30m/s2 divided by 2 = 2700 j

3) how can i find the energy lost do i subtract the kinectic with the potential

4) how can i explain how the energy was lost was it because it was falling?

The Attempt at a Solution

is their a formula to finding mechanical energy lost?

 

[xcvxcvvc]

Homework Statement

a 6.0kg concrete block is dropped from the top of a tall building. the block has fallen a distance of 55 meters and has a speed of 30 meters per srcond when it hits the ground

1) at the instant the block was released what was the gravitational potenetial energy with respect to the ground?

2)calculate the kinetic energy of the block at the point of impact?

3) how much mechanical energy was lost by the block as it fell?

4) explain what happened to the mechanical energy that was lost by the block?

Homework Equations

1) pe = mgh so 6kg x 9.81m/s2 x 55m = 323 j

2) ke = 1/2mv2 so 6kg x 30m/s2 divided by 2 = 2700 j

3) how can i find the energy lost do i subtract the kinectic with the potential

4) how can i explain how the energy was lost was it because it was falling?

The Attempt at a Solution

is their a formula to finding mechanical energy lost?

First off, for part 1.) you lost a zero. The answer is 3.23 KJ = 3230 J. You should have spotted this due to the unrealistic nature of an object gaining more kinetic energy than was available in potential energy. Second, for part 3, you have it reversed. The total energy available at t = 0 is the potential energy. By the time it hits the ground, all potential energy is gone. Some of it is kinetic and some of it is that sought after mechanical loss. I'd explain the losses with friction of the air -- also known as wind resistance.

 

[zelda1850]

so the mechanical energy was lost due to friction of the air?