# Questions regarding conservation of energy with heat

1. Nov 15, 2007

### VanKwisH

1. The problem statement, all variables and given/known data
Calculate the mechincal energy convertered to thermal energy when a 8.0kg
box is pushed 5.0 m along a 30 degree cinline at a constant velocity by an
average force of 75 N parallel to the incline.
Hint : how much work is done in pushing the box along the incline?
What is the Gravitational Potential Energy of the box when it is at the top of the incline?
why are these value's not the same?

2. Relevant equations
Pe1 + Ke1 = Pe2 + Ke2 + Heat
mgh1 + 1/2 mv1^2 = mgh2 + 1/2mv2^2 + (FF* D)

3. The attempt at a solution

I started by calculating the height of the incline which is
H = 5 sin (30) = 2.5m
Pe2 = 8 * 9.8 * 2.5 = 196J

but now i am completly stuck ..............

Last edited: Nov 15, 2007
2. Nov 15, 2007

### hotcommodity

If the "1's" and "2's" in your equation are supposed to represent final and inital respectively, I would suggest placing the "heat" on the other side of your equation. The heat is the work done by a non-conservative force. The work done by a non-conservative force is given by $$W_{nc} = E_f - E_0$$.

Show me where you're plugging in numbers, and maybe we can find the problem.

Edit: If you have a new problem, I would suggest making a separate thread for it rather than editing your original post. It can be confusing for anyone trying to help you out.

Last edited: Nov 15, 2007
3. Nov 15, 2007

### VanKwisH

yah that's kinda the problem i have i don't exactly know what to plug in but i have plugged in .... and yes the 1's and 2's represent the initials and finals

H = 2.5
mgh2 = 8*98*5
heat = ff * d
75 * 5 = 375 J

mgh1 + 1/2mv2^2 = (8*9.8*2.5) + 1/2mv2^2 + ( 375 )

that's what i have so far ...........
i know when i do 375 - ( 8*9.8*2.5 ) i get 179J
but i can't explain the math/physics involved

4. Nov 15, 2007

### hotcommodity

You get 179 Joules, but you only know each value to two significant digits, and thus the answer becomes 180 Joules. I assume you're stuck on the part about why the work done on the box is not the same as the potential energy of the box at the top of the incline. The mechanical energy of the box is only conserved if what does not do negative work on the box? This should tell you why the two values are not the same.

5. Nov 15, 2007

### VanKwisH

so if gravity doesn't do work on the box ??? then the energy conserved????

6. Nov 15, 2007

### hotcommodity

The gravitational force is a conservative force. If the crate slides back down the incline, and we and ignore friction, then mechanical energy is conserved, in other words, the amount of potential energy that the box had at the top of the incline would be completely transfered to kinetic energy right before it hit the very bottom of the incline. Total mechanical energy is not conserverved when forces like friction act on the object in question. Friction is a non-conservative force, and it causes objects to lose mechanical energy (give off heat). Look at the equation in the second post, and you'll see why this is mathematically true. Does that help?

7. Nov 15, 2007

### VanKwisH

i can somewhat understand but
what does the equation in the 2nd post mean exactly... in words...
i don't think i use the same variables as u do ........

8. Nov 15, 2007

### hotcommodity

It means that the net work done by all of the non-conservative forces, not to be confused with the net work done on the object, is equal to the change in the objects mechanical energy. If Wnc was zero, then I'd know that mechanical energy is conserved, Ef = E0. So we saw that the friction, a non-conservative force, did negative work on the box. This can be found by observing the change in mechanical energy of the box over the 5 meters. The equation in post two didn't help us find the work done by friction, but it helps to explain why the potential energy of the box is not equal to the initial kinetic energy of the box. Does that help?

9. Nov 15, 2007

### VanKwisH

oooooooooo yah i got it now ........ thnx a lot