Work done by Friction on an Inclined Plane

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SUMMARY

The discussion focuses on calculating the work done by friction on a suitcase being pulled up an inclined plane. The suitcase has a mass of 17.1 kg, is pulled by a force of 146 N at an angle of 24.0 degrees, and experiences a coefficient of kinetic friction of 0.257. The initial calculation of the friction force was incorrectly assumed to be equal to the product of the coefficient of friction and the gravitational force. The correct approach involves recognizing that the normal force is influenced by both gravity and the incline's angle, leading to a revised friction force calculation of 43.06806 N and a work done by friction of -168 J.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with inclined plane physics
  • Knowledge of friction coefficients and their application
  • Ability to construct and analyze free body diagrams
NEXT STEPS
  • Study the concept of normal force on inclined planes
  • Learn about the work-energy principle in physics
  • Explore the effects of different coefficients of friction on motion
  • Practice solving problems involving forces on inclined planes
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and inclined planes, as well as educators seeking to clarify concepts related to friction and work done by forces.

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Homework Statement



A luggage handler pulls a suitcase of mass 17.1 kg up a ramp inclined at an angle 24.0 above the horizontal by a force of magnitude 146 N that acts parallel to the ramp. The coefficient of kinetic friction between the ramp and the incline is 0.257. The suitcase travels a distance 3.90 m along the ramp.

What is the work done on the suitcase by the friction force?

Homework Equations



W = F o r = |F|*|r|*cos(theta)
Friction Force = uN

The Attempt at a Solution



Ok so I calculated that the Friction Force should equal umg since the direction of movement is along the incline rather than in the x direction. So Friction Force = 43.06806 N. If we multiply it by the change in distance, the Work done by friction should be -168 J, but this doesn't seem to be the correct answer. Anyone know where I may have screwed up?
 
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well force of friction is not \mu mg . set up a free body diagram first.
 
Oh wow... I'm sorry that was a stupid move on my part. For some reason I forgot that the normal force depends on gravity AND the direction of the gravitational force... Thanks so much I appreciate it!
 

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