# finding mechanical energy lost

Discussion in 'Physics' started by zelda1850, Feb 21, 2010.

1. ### zelda1850 Thread Starter New Member

Jan 3, 2010
18
0
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?

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

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

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

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

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

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

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

2. ### mik3 Senior Member

Feb 4, 2008
4,846
63
Mechanical energy=Potential energy + Kinetic energy

At the instant of the impact, potential energy =0 and thus mechanical energy=kinetic energy. The mechanical energy loss is equal to the kinetic energy at the impact because after the impact the block has no kinetic neither potential energy.

The mechanical energy lost was converted to heat, acoustic energy (he sound of the impact) and the energy required to deform the block.

3. ### zelda1850 Thread Starter New Member

Jan 3, 2010
18
0
does that mean kinetic is equal to the gravational ?

4. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Not exactly. Kinetic is the speed at which something is moving. Gravitational is the potential energy it has, which won't be realized until you let it fall.

Air friction will be a major cause of loss. All things have a terminal velocity, where the speed reaches a maximum and can't increase any more due to air friction.

In the case of meteorites, which are falling much faster than terminal velocity, the immediate result is heat, lots of it. The meteor is also slowing down very quickly, which is how the various space craft over the years do it, atmospheric braking.

5. ### mik3 Senior Member

Feb 4, 2008
4,846
63
Assuming no friction, it will be equal at the instant of the impact.

6. ### BillO Distinguished Member

Nov 24, 2008
985
136
There is something wrong with your arithmetic.

1) PE between start and point of impact = 6x9.81x55 = 3237j
2) KE at point of impact = 6*30*30/2 = 2700j
3) Energy lost during fall = 3237-2700 = 537j
4) Lost to heat due to wind resistance

Edit: This can be verified by doing the calculations for a 55m drop in a vacuum which give a terminal velocity of a bit less than 3.35m/s and a KE at the point of impact of just less than 3240j, which is within the margin of error for the precision given.

Last edited: Feb 23, 2010