Physics Machine efficiency question.
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Physics Machine efficiency question.

[From: ] [author: ] [Date: 11-06-19] [Hit: ]
mass a distance of 5.65 meters using a machine 72.5 percent efficient?If the answer is 16420.07 N, Im right!......
Hi, I want to make sure I understand the concept of efficiency if you guys get the same answer as me.

What work is required to lift a 215 kg. mass a distance of 5.65 meters using a machine 72.5 percent efficient?

If the answer is 16420.07 N, I'm right! If it's not, and it probably will be wrong, can you please show me your work?

Thanks in advance =)

W = m*g*D/η = 215*9.8*5.65/.725 = 16420.07 J (not Newtons, a measure of force)

Keep up the good work!

The work done in lifting the object is equal to it's change in gravitational potentional energy. However since the machine is only 72.5% efficient this means that only 72.5% of the work it does goes into changing the gravitational potential energy of the mass.


72.5/100 * Work done by machine = m*g*h
= 215*9.8*5.65
= 11,904.55 J
Work done by machine = 11,904.55/(72.5/100)
= 16,420.07 J (!!!)

Your working is correct and you have the correct answer as far as the number is concerned. However, the unit is Joules (J) not N (newtons). Work required to lift is the energy given to the mass when it is lifted not the force applied to the mass as it is lifted.
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