Efficiency and Energy to move a box on a conveyor belt

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SUMMARY

The discussion focuses on calculating the force exerted on a box by a conveyor belt and the work done in moving the box over a distance. The box has a mass of 35.7 kg, and using Newton's second law (F=ma), the force exerted by the belt is calculated as 99.2 N. The work done in moving the box 100 meters is determined using the formula W=fd, resulting in 9920 J. The final energy input required, considering an efficiency of 25%, is calculated to be 3.97 x 10^4 J.

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  • Understanding of Newton's second law (F=ma)
  • Knowledge of work-energy principle (W=fd)
  • Basic concepts of force and mass
  • Familiarity with efficiency calculations
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UnknownQuestioner
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Homework Statement
Please refer to the attached picture.
Relevant Equations
This is where I am stuck at, I can’t seem to figure out the formulas to use and the combination of using them.
BNuT1m6SNQp9NkuEX_uqpdE9Mzaiexp13KOhfix3YyAE3Q5Y5n9hPU6QqUzu2HeNAKFm5-L_5OpYv8QI0qRLzm_iCbtBNS...png
 
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What force acts on the box? I assume the conveyor belt is horizontal.
 
Gravity? Not sure
 
If the box is being accelerated by the belt, the belt's got to be exerting a force on it. You know the mass of the box (from the weight) and the acceleration, so can you get the force the belt exerts on the box?
 
1588848049429.png
 
Yeah, but you can just use components here. Just say ##F=ma##. What is the force?
 
Would that be 350 x 2.78 to get 973 N as the force?
 
UnknownQuestioner said:
Would that be 350 x 2.78 to get 973 N as the force?

What's the mass of the box?
 
Oops, that would be 35.7 kg. Then 35.7 x 2.78 to get 99.2 N
 
  • #10
UnknownQuestioner said:
Oops, that would be 35.7 kg. Then 35.7 x 2.78 to get 99.2 N
So how much work is done on the box in moving it the 100m?
 
  • #11
haruspex said:
So how much work is done on the box in moving it the 100m?

I used f=ma to get 99.2N then I used W=fd to get 9920J. This energy is the output, so output/0.25=input. To get a final answer of 3.97 x 10^4 J This is correct?
 
  • #12
UnknownQuestioner said:
I used f=ma to get 99.2N then I used W=fd to get 9920J. This energy is the output, so output/0.25=input. To get a final answer of 3.97 x 10^4 J This is correct?
Looks good.
 

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