A Mover Pushes a Piano Up a Ramp: Work Done?

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Homework Help Overview

The discussion revolves around a scenario where a mover pushes a piano up a ramp, raising it to a stage. The problem involves calculating the work done by the mover and the work done by the machine, which is interpreted as the ramp in this context. The subject area includes concepts of work, force, and mechanical advantage in physics.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of work done by the mover using the formula W=F dot x and question the definition of "work done by the machine." There is exploration of the mechanical advantage provided by the ramp and considerations of friction affecting the work done.

Discussion Status

Some participants have provided calculations for the work done by the mover and attempted to clarify the concept of work done by the machine. There is ongoing uncertainty regarding the interpretation of the term "machine" and how it relates to the work done, particularly in the context of friction.

Contextual Notes

Participants are working with specific values such as the weight of the piano and the dimensions of the ramp. There is mention of friction affecting the calculations, and the discussion reflects a need for clarity on the definitions and assumptions involved in the problem.

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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...
 
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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?
 
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.
 
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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