The incredible lifting machine heat loss?

tubworld
Messages
38
Reaction score
0
the incredible lifting machine heat loss??

While stumbling upon this website on the Incredible lifting machine,

http://tigre.ucr.edu/dipen/phy11/heat.htm
I was wondering what could be some of the reasons why the work down calculated as in mgh does not equal to the work done under the pv graph?

I thought it should be the same? if not, that means there is heat loss by via what means?
 
Last edited by a moderator:
Physics news on Phys.org
tubworld said:
While stumbling upon this website on the Incredible lifting machine,
http://tigre.ucr.edu/dipen/phy11/heat.htm
I was wondering what could be some of the reasons why the work down calculated as in mgh does not equal to the work done under the pv graph?
I thought it should be the same? if not, that means there is heat loss by via what means?
What suggests to you that the increase in potential mechanical energy of the system is not equal to \int Pdv?

AM
 
Last edited by a moderator:
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
Back
Top