Calculate Efficiency of a Falling Mass System: Homework Help and Solution - 82%

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The discussion focuses on calculating the efficiency of a system where a falling mass moves a box along a desk. The efficiency is calculated using the formula Efficiency = (energy output/energy input) x 100%, with the energy input and output derived from the weights and distances involved. Initial calculations yielded an efficiency of 82%, but there were concerns about the formulas used for energy input and output being reversed. The conversation highlights that friction affects efficiency and that the gravitational potential energy lost by the falling mass must equal the kinetic energy gained by the moving mass. Ultimately, the efficiency calculation must account for the actual forces acting on the sliding mass, indicating it will always be less than 100%.
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Homework Statement


Question:
Calculate the efficiency of a system that uses a falling mass to move a box along the desk.
Given (helps to see picture):
Weight of (sliding) box = 164g
Weight of (falling) mass = 200g
Distance of board (platform) = 88cm
Time (avg) = 6s
(anything missing? I can get it, just point it out [its a lab])

Homework Equations


Efficiency = (energy output/energy input) x 100%

The Attempt at a Solution


Ei = mad
= (0.2kg)(9.8kg/N)(0.88m)
= 1.7J
Eo= mgh
= (0.164kg)(9.8kg/N)(0.88m)
= 1.4J
E% = (Eo/Ei) x 100%
= (1.4J/1.7J) x 100%
= 82%

Picture I made:

w.w.w.i235.photobucket.com/albums/ee170/ShadowedIMG/Homework.png
(cant post links)

Any input to anything would be very helpful, and thanks for your time.
 
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You need to know friction. Without friction, the efficiency is 100%.

UPDATE: No, never mind. That's wrong. What you have is right, except that the formulas for Ei and Eo are reversed. (but friction will decrease the efficiency)
 
Last edited:
well this problem sort of seems to be solved wrong because of the simple fact the effeciency of the system is merely a factor of \frac{E_i}{E_o} By your calculations, if you go to see, acceleration, the distance moved all cancel out and hence the effeciency is given by the ratios of the two masses. Hence, by just adjusting the masses of the two systems, you can raise the effeciency of the system to more than 100%.

although i am not sure how this contradiction occurs, and the formula which is W = Fd seems to be used rightly.. the loss in the GPE of the falling mass MUST equal the kinetic energy acquired by the moving mass.
 
Hi UnnamedOne,

UnnamedOne said:

Homework Statement


Question:
Calculate the efficiency of a system that uses a falling mass to move a box along the desk.
Given (helps to see picture):
Weight of (sliding) box = 164g
Weight of (falling) mass = 200g
Distance of board (platform) = 88cm
Time (avg) = 6s
(anything missing? I can get it, just point it out [its a lab])

Homework Equations


Efficiency = (energy output/energy input) x 100%


The Attempt at a Solution


Ei = mad
= (0.2kg)(9.8kg/N)(0.88m)
= 1.7J
Eo= mgh
= (0.164kg)(9.8kg/N)(0.88m)
= 1.4J

I think you have a problem with a value here. (Your units of kg/N need to be N/kg.) Using 9.8 N/kg is correct for the g in the energy input mgh; however, 9.8 is not correct for the acceleration of the sliding mass. It will always be less than 9.8 (even if there is no friction).

That's why the efficiency here will always be less than 100% even if there is no friction; gravity is doing work on both blocks, but you only want to find the fraction of gravity's work that is going to the sliding block. (Using mad for the energy output would include frictional heating as part of the energy output.)
 
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