Total Mechanical Energy

[sbook2]

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I did a Kaplan discrete question:

A 40 kg block is resting at a height of 5m off the ground dropped. If the block is released and falls to the ground, which of the following is closest to its total mechanical energy at a height of 2m, assuming negligible air resistance?

So, I solved the question using Total Mechanical Energy = Potential Energy + Kinetic Energy (E= U+K)

I did E= (40)(10)(2) + (1/2)(40)(10^2), which gives 2800 J, however the correct answer given is 2000 J. They didn't include KE in the calculation and I don't understand why. The explanation is that the kinetic energy at the starting point is 0, but it has KE as it's falling so it has KE at 2m, which is why I thought it would be included in the calculation. Anyone know why the don't include KE?

 

[kadler538]

So, the question is really trying to show that total mechanical energy is conserved throughout the fall (assuming no air resistance). The total mechanical energy that the block starts with on the table is the total mech E that it will end with when it hits the ground, because energy is conserved. Yes, it does lose PE and gain KE as it falls, however the LOSS in PE will be translated into the GAIN in KE. So the PE and KE DO change, but at the end, their sum will still equal the Total Mechanical Energy that the block had at the START, which is mgh which = (40kg)(10m/s^2)(5m) = 2000 J 🙂 That's what the question is interested in.

 

[sbook2]

So, the question is really trying to show that total mechanical energy is conserved throughout the fall (assuming no air resistance). The total mechanical energy that the block starts with on the table is the total mech E that it will end with when it hits the ground, because energy is conserved. Yes, it does lose PE and gain KE as it falls, however the LOSS in PE will be translated into the GAIN in KE. So the PE and KE DO change, but at the end, their sum will still equal the Total Mechanical Energy that the block had at the START, which is mgh which = (40kg)(10m/s^2)(5m) = 2000 J 🙂 That's what the question is interested in.

Thanks 🙂

 

[aldol16]

Above explanation is correct - however, you should still get the same answer the way you did it. In fact, your kinetic energy should be 1200 J. How did you get your velocity?

 

[kadler538]

Above explanation is correct - however, you should still get the same answer the way you did it. In fact, your kinetic energy should be 1200 J. How did you get your velocity?

agreed 🙂 looks like the OP used acceleration rather than final velocity?

to clear this up, if you used vf^2 = vo^2 + 2ax then you should get a final velocity of vf^2 = 60 (remember that x = 3m because that is how far the block has fallen) which would make the Total Mechanical Energy = (40kg)(10 m/s^2)(2m) + 1/2(40kg)(60) = 2000 J.