Power Developed for Conveyor Belt Transporting Newspapers

  • Thread starter Thread starter Lothar
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AI Thread Summary
The discussion focuses on calculating the power developed by a conveyor belt transporting bundles of newspapers. The conveyor is 13.5 meters long and inclined at 30 degrees, moving 15 bundles per minute, with each bundle containing 26 newspapers weighing 1 kg each. The key equation for work is ΔPE = m*g*Δh, where the height change is calculated using the sine of the incline. Participants suggest calculating the total work done in one minute to simplify finding the power output. The correct approach involves determining the work done for all newspapers and converting it to power in watts.
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Homework Statement


A 13.5 m long conveyor belt, inclined at 30.0°, is used to transport bundles of newspapers from the mailroom up to the cargo bay to be loaded on to delivery trucks. Each newspaper has a mass of 1.0 kg, and there are 26 newspapers per bundle. Determine the power that the conveyor develops if it delivers 15 bundles per minute.

Homework Equations


w = fd
w = fdcos(theta)
P = w/t

The Attempt at a Solution


I've tried several things getting incorrect answers of:
57.33
859.95
744.74
49.6
372.4
 
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Lothar said:

Homework Statement


A 13.5 m long conveyor belt, inclined at 30.0°, is used to transport bundles of newspapers from the mailroom up to the cargo bay to be loaded on to delivery trucks. Each newspaper has a mass of 1.0 kg, and there are 26 newspapers per bundle. Determine the power that the conveyor develops if it delivers 15 bundles per minute.

Homework Equations


w = fd
w = fdcos(theta)
P = w/t

The Attempt at a Solution


I've tried several things getting incorrect answers of:
57.33
859.95
744.74
49.6
372.4

Welcome to PF.

Maybe focus on your last equation? If you figure the work performed in 1 minute you're just about home free right?

So what is the work of moving 1 newspaper to the top of the ramp?

W = ΔPE = m*g*Δh

where Δh = 13.8*sin30 = 1/2*13.8 = 7.9 m

Figure then the total work from 15*26 newspapers and that tells you how much for a minute, so getting to work/sec is really easy then.
 
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