Questions regarding conservation of energy with heat

In summary, to calculate the mechanical energy converted to thermal energy when pushing a 8.0kg box along a 30 degree incline at a constant velocity, use the equation mgh1 + 1/2mv1^2 = mgh2 + 1/2mv2^2 + (FF*D), where h is the height of the incline and FF is the force of friction. The values for gravitational potential energy and work done by pushing the box are not the same because of the work done by non-conservative forces, such as friction, which causes the object to lose mechanical energy.
  • #1
VanKwisH
108
0

Homework Statement


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?

Homework Equations


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

The Attempt at a Solution



the answer is 180 J
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 completely stuck .....
 
Last edited:
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  • #2
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 [tex] W_{nc} = E_f - E_0 [/tex].

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.
 
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  • #3
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
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
so if gravity doesn't do work on the box ? then the energy conserved?
 
  • #6
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 transferred 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
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
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
oooooooooo yah i got it now ... thnx a lot
 

1. What is the conservation of energy with heat?

The conservation of energy with heat is a fundamental law of physics that states that energy cannot be created or destroyed, but can only be converted from one form to another. This means that the total amount of energy in a closed system remains constant, even as it changes form.

2. How does heat relate to the conservation of energy?

Heat is a form of energy that is transferred between objects or systems that are at different temperatures. The conservation of energy requires that the total amount of heat transferred must equal the total change in energy of the system.

3. What are some examples of the conservation of energy with heat?

Some examples of the conservation of energy with heat include a pot of water boiling on a stove, a car engine converting fuel into heat and motion, and a light bulb producing heat and light from electrical energy.

4. How does the conservation of energy with heat impact our daily lives?

The conservation of energy with heat has a significant impact on our daily lives as it governs the behavior of many natural and man-made systems. It allows us to understand and control various processes such as cooking, heating and cooling our homes, and producing electricity.

5. Are there any exceptions to the conservation of energy with heat?

While the conservation of energy is a fundamental law, there are some situations where it may seem like energy is not conserved. For example, when heat is transferred from a warmer object to a cooler object, the total energy of the system remains constant, but the energy is redistributed. This redistribution can make it seem like energy is lost, but it is simply being converted into a different form.

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