Simple Question?

  • Thread starter bbg5000
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  • #1
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Simple Question..??

A mover pushes a 260-kg piano on wheels up a ramp 7.0 m long onto a stage 1.75 m above the auditorium floor. The mover pushes the piano with a force of 680-N
a) How much work is done by the mover?
b) What is the work done on the piano by the machine?

Help...
 

Answers and Replies

  • #2
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A) W=F dot x... ramp is 7m long and the mover is pushing in the same direction as the ramp with 680N. W= 680N * 7.0m = 4800j.

B) What is the machine?
 
  • #3
HallsofIvy
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Assuming the "machine" is the ramp (and not the mover!):

The force necessary to "lift" a 260 kg piano is its weight- 260*9.8= 2450 Newtons.
The "mechanical advantage" due to the ramp is 2450/680= 3.6, slightly less than the ratio of the height of the ramp to its length- apparently there is a little friction involved.

I'm still not sure what is meant by "work done by the machine". The work done in lifting a 260 kg mass 1.75m is its weight time distance: (2450)(1.75)= 4287.5 Joules.
That may be what is intended.

The actual work done by the mover is the force applied times the distance moved:
680*7= 4760 Joules. That additional 4760- 4287.5= 472.5 Joules work is due to that friction I mentioned.
 
  • #4
Doc Al
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HallsofIvy said:
Assuming the "machine" is the ramp (and not the mover!):

... I'm still not sure what is meant by "work done by the machine".
Assuming, as you did, that the machine is the ramp, then I would say that the work done by the machine is that due to friction (negative work). The ramp exerts two components of force on the piano, but the normal component does no work, since the displacement is zero in that direction.
 

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