Percentage energy loss (mechanical energy) problem

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SUMMARY

The discussion centers on calculating the percentage energy loss of a 2.0 kg object sliding down a low-friction incline, reaching a speed of 2.7 m/s at the bottom from a height of 0.50 m. The correct approach involves using the energy conservation equation: mgh(0) + W(other forces) = 0.5mv². Participants emphasize the importance of isolating W(other forces) by subtracting the initial potential energy (mgh(0)) from the final kinetic energy (0.5mv²) before calculating the percentage loss. A common mistake noted is the failure to multiply the result by 100 to convert it to a percentage.

PREREQUISITES
  • Understanding of gravitational potential energy (mgh)
  • Familiarity with kinetic energy (0.5mv²)
  • Basic algebra for manipulating equations
  • Knowledge of percentage calculations
NEXT STEPS
  • Review energy conservation principles in physics
  • Practice problems involving percentage calculations in energy loss
  • Learn about the effects of friction on energy loss in mechanical systems
  • Explore advanced topics in mechanics, such as work-energy theorem
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators seeking to clarify concepts related to energy loss in systems.

poopandpee
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Homework Statement


Problem: An object with mass 2.0 kg slides down a low-friction incline and its speed measured at the bottom is 2.7 m/s. The object starts sliding at a height of 0.50 m over the tabletop and its speed is measured 0.10 m over the tabletop. How big is the percentage energy loss?

Homework Equations

The Attempt at a Solution


My attempt: I've tried using this formula: mgh(0) + W(other forces) = 0.5mv2, where m=2.0 kg, h=0.4 m, and v=2.7 Then I found W(other forces) and divided it by mgh(0). However, that did not give a right answer. Any suggestions?
 
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The sign of the energy loss is a bit questionable but overall the approach is good. Maybe some mistake in the calculations? You didn't show them.
 
Did you remember to multiply by 100 to convert to %?
 
You shouldn't divide by the other work, if your equation above is correct you should subtract mgh(0) to get W alone
 
NateTheGreatt77 said:
You shouldn't divide by the other work, if your equation above is correct you should subtract mgh(0) to get W alone
That appears to be what @poopandpee did before dividing:
poopandpee said:
Then I found W(other forces)
As others have posted, the method sounds correct, but we need to see the details.
 

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