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

  • Thread starter Thread starter bbg5000
  • Start date Start date
AI Thread Summary
A mover pushes a 260-kg piano up a 7.0 m ramp with a force of 680 N. The work done by the mover is calculated as 680 N multiplied by 7.0 m, resulting in 4760 Joules. The weight of the piano requires a lifting force of 2450 N, leading to a calculated work of 4287.5 Joules for the ramp. The difference of 472.5 Joules indicates the work done against friction. Clarification on the term "work done by the machine" suggests it refers to the ramp's contribution, particularly the friction involved.
bbg5000
Messages
14
Reaction score
0
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...
 
Physics news on Phys.org
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.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Back
Top