How much heat is generated from lost mechanical energy?

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Homework Help Overview

The discussion revolves around calculating the heat generated from lost mechanical energy in a physics problem involving an object in motion, friction, and energy conversion. The context includes the mechanical equivalent of heat and the relationship between Joules and calories.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between mechanical energy loss and heat generation, questioning the units of measurement and the implications of given values. Some participants attempt to derive stopping distance and time, while others clarify the conversion of energy units.

Discussion Status

The discussion is active, with participants providing hints and guidance on how to approach the problem. There is an ongoing exploration of the calculations needed to determine the generated heat quantity, and some participants have clarified the meaning of the mechanical equivalent of heat.

Contextual Notes

There are concerns about the clarity of the problem statement, particularly regarding the units of measurement and the assumptions made about energy conversion percentages. Participants are navigating these ambiguities while attempting to solve the problem.

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Let the mechanical equivalent of heat be 4.2[J/cal]. 70 percent of lost mechanical energy was converted into heat. Find the generated heat quantity given an object with mass 10 kg moving at speed 49 m/s with coefficient of kinematic friction 0.5 and gravitational acceleration of 9.8 [tex]\frac{m}{s^2}[/tex].
 
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willydavidjr said:
Let the mechanical equivalent of heat be 4.2[J/cal]

Joules per calorie?
 
Yes Hootenanny! I really cannot get it because the givens are quite unique and confusing.
 
I think there's a typo in the question because the units of J/cal are nonsensical. Heat is 'energy in transit' therefore the SI units are Joules.
Regards,
~Hoot
 
Last edited:
I am trying to solve the problem but all I can get is the stopping distance. I am still researching for the formula for the time respected to the problem.
 
Nevermind, I have seen my mistake. If you have calculated the stopping distance you have all the information you need.

HINT: What is work done?

Regards,
~Hoot
 
Can I use the formula [tex]x-x_o=v_ot + \frac{1}{2}at^2[/tex] to get the stopping time?Or is there any other formula?
 
If you have calculated the stopping distance, you have all the information you need. Work done = force times distance. In this case work done = energy 'lost'.

Regards,
~Hoot
 
If I get the work with this= 0.5 * 10 kg * 9.8 [tex]\frac{m}{s^2}[/tex] * 255 m, then what is the purpose of the given "mechanical equivalent of heat be 4.2[J/cal]. 70 percent of lost mechanical energy was converted into heat"?
 
  • #10
You should get 245m, but 255 is close enough. You have calculated the energy lost. However, only 70% of this 'lost' energy is converted into heat...

Regards,
~Hoot
 
  • #11
Yeah you're really right. That was really close enough for the stopping distance. But how can I compute for the generated heat quantity given the mechanical equivalent of heat be 4.2[J/cal]. The final answer is suppose to be in calorie unit.
 
  • #12
Ahh, I think what they question meant was 1 calorie = 4.2 Joules.

Regards,
~Hoot
 
  • #13
Now I get it buddy. 12495 Joules is the work done or the energy lost and I will multiply it by 0.70 to get the 70 percent lost. I get 8746.5 Joules. Dividing it by 4.2 j/cal gives me 2082.5 cal. Is 2082.5 right buddy?
 
  • #14
Yep, that looks good to me.

Regards,
~Hoot
 
  • #15
Thanks Hoot. You were a great help to me. Till next time.
 

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