# Physics 11 kinetic and potential energy2

1. Dec 20, 2006

### Iceclover

1. The problem statement, all variables and given/known data

A 98N sack of grain is hoisted to a storage room 50m above the ground floor of a grain elevator.
a) how much work was required
b) what is the potential energy of the sack of grain at this height
c) the rope being used to lift the sack of grain breaks just as the sack reaches the storage room. what kinetic energy does the sack have just before it strikes the ground floor?

2. Relevant equations

3. The attempt at a solution

i got the same answer each time. I got 4.9x10^3 for a, b and c

2. Dec 20, 2006

### mbrmbrg

Me too. Do you know why they're all the same?

3. Dec 20, 2006

### Iceclover

the work-energy therom?

4. Dec 20, 2006

### Iceclover

and if an object was falling and was losing 400J of potential energy how much kinetic energy would it gain?

5. Dec 20, 2006

### mbrmbrg

funny dude.

OK, for each specific case:
Why does energy needed to lift=potential energy at top?
More basically, why does potential energy at top=kinetic energy at bottom?

6. Dec 20, 2006

### mbrmbrg

No, no. You tell me. This be basic, basic theorem. Veeeerrrrrrry important. And it is not the Work-Energy Theorem (though \it could be related to it rather neatly).

7. Dec 20, 2006

### Iceclover

would it gain 400J? i really just don't get this

8. Dec 20, 2006

### mbrmbrg

Yes it would be 400J. There's not much to get. The conservation of energy states quite simply that under certain conditions, mechanical energy remains constant. This is written up as $$E_{mec}=K+U=constant$$ (K is kinetic energy and U is potential energy).

So any loss in potential energy requires an equal gain in kinetic energy, and vice-versa.

The usual analogy is of two glasses and a given amount of liquid. If you pour the liquid from one glass to the other (without spilling), whatever amount one glass loses, the other glass gains.