Physics problem (Work/Power)?

👨🏻‍🏫 學術台物理學

[匿名]

2009-04-14 3:19 am

I've tried this problem several times, yet I keep getting the answer wrong. I'm probably making some small mistake somewhere, but I can't seem to find it.

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.

更新1:

@ Lucifer I've tried that and it comes out to 57.33, which is not the right answer.

回答 (5)

[匿名]

2009-04-14 3:44 am

✔ 最佳答案

P=work/time
P=Fd/time
P=mgd/time
mass = (1)(26)(15) = 390 kg
g = 9.8 m/s^2
time = 60 s
i think you may have messed up on the distance. it should be the length of the belt times sin(30) because that is the distance it is lifted vertically.
d=13.5sin(30)=6.75 m

P=(390)(9.8)(6.75)/60
so I believe the power should be

P=430. watts

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[匿名]

2009-04-14 10:47 am

Assuming
1) the weight of belt and relevant components are not significant.
2) no slip between the newspaper bundle and the belt surface.
3) always 15 bundles on the belt.
4) g = 9.81ms^-2
Total Power:
= Power to bring the bundles by horizontal distance of 13.5cos30Âº + Power to lift the bundles up vertically by height of 13.5sin30Âº
= (26x1) x g x 13.5 x (cos30Âº + sin30Âº) x15 / 60
= 745.5W + 430.4W ------------ (horizontal + vertical displacements)
= 1176W
~ 1.18kW

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Steve

2009-04-14 10:40 am

mass delivered = 26 kg * 15 bundles = 390 kg
height increase = 13.5 m *sin30Â° = 6.75 m

Energy = mg*h = 390*9.8*6.75 = 25798.5 J

P = E/t = 25798 J/60 sec = 430 watts

Since nothing was said about friction, I had to assume it is not considered here. In real life, it would be a major source of power requirement..........

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[匿名]

2009-04-14 10:37 am

The work done on the bundles is m*g*âh, where m = total mass of papers, âh is the change in elevation. In this case, âh = L*sinÎ¸

âE = 26.0*15*9,8*13.5*sin30Âº = 25800 J

This is over one min, or 60 sec, so the power is 430 W

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[匿名]

2009-04-14 10:24 am

Power = work / Time

Work = Force * Distance

In this case, distance is your conveyor belt length, force is the weight of your newspapers

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