# Generated Heat Quantity

1. Apr 17, 2006

### willydavidjr

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 $$\frac{m}{s^2}$$.

2. Apr 17, 2006

### Hootenanny

Staff Emeritus
Joules per calorie?

3. Apr 17, 2006

### willydavidjr

Yes Hootenanny! I really cannot get it because the givens are quite unique and confusing.

4. Apr 18, 2006

### Hootenanny

Staff Emeritus
I think theres 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: Apr 18, 2006
5. Apr 18, 2006

### willydavidjr

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.

6. Apr 18, 2006

### Hootenanny

Staff Emeritus
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

7. Apr 18, 2006

### willydavidjr

Can I use the formula $$x-x_o=v_ot + \frac{1}{2}at^2$$ to get the stopping time?Or is there any other formula?

8. Apr 18, 2006

### Hootenanny

Staff Emeritus
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

9. Apr 18, 2006

### willydavidjr

If I get the work with this= 0.5 * 10 kg * 9.8 $$\frac{m}{s^2}$$ * 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. Apr 18, 2006

### Hootenanny

Staff Emeritus
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. Apr 18, 2006

### willydavidjr

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. Apr 18, 2006

### Hootenanny

Staff Emeritus
Ahh, I think what they question meant was 1 calorie = 4.2 Joules.

Regards,
~Hoot

13. Apr 18, 2006

### willydavidjr

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. Apr 18, 2006

### Hootenanny

Staff Emeritus
Yep, that looks good to me.

Regards,
~Hoot

15. Apr 18, 2006

### willydavidjr

Thanks Hoot. You were a great help to me. Till next time.