Efficiency of Ramp: Solving Box Push Up Problem

AI Thread Summary
The discussion focuses on calculating the efficiency of a ramp used to push a 7.0 kg box up a height of 6 m with a hypotenuse of 12 m in 3.25 seconds. The force required to push the box at a constant velocity is 40 N, which is considered the input force (Win). The output work (Wout) is calculated using the normal force, determined to be 68.6 N, resulting in an output work of 823 J. The efficiency is then derived from the ratio of output energy to input energy, leading to the conclusion that the efficiency is based on how much energy is effectively transferred to the box compared to the energy expended. The discussion emphasizes the importance of correctly identifying the forces involved in the calculation to determine the ramp's efficiency accurately.
Zack K
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Homework Statement


A 7.0 kg box is pushed up the ramp (height is 6 m and hypo is 12 m) in 3.25 s. If it requires a force of 40 N to push at a constant velocity, what is the efficiency of the ramp?

Homework Equations


Eff=(Wout/Win) x100
W= Fxd
F=ma

The Attempt at a Solution


The problem I have is that I don't know if 40.0 N is the force for Wout or Win. But I am guessing that it is Win since it's the minimum force need to push at a constant velocity. Then to get Wout you find the Normal force of the box which is 7 kg x 9.8= 68.6 N then you use W=68.6 x 12= 823 J. 480/823=58% efficient. But of course that's not right.
 
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How much energy goes into the box pushing it up the ramp?
How much of that energy does the box finish up with?
 
Zack K said:
Normal force of the box which is 7 kg x 9.8= 68.6 N
since it's a ramp, the normal force would not equal to the weight of object.
 
So by definition the efficiency is eff=(how much energy you got out)/(how much energy you put in). In this problem, the 40N is the force you put in and the potential energy is the energy you get out. Hope that helps you
 
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