Work done by friction and change in mechanical energy

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SUMMARY

The discussion focuses on the mechanics of a 12-kg crate sliding up a 2.5-meter ramp, emphasizing the role of friction in determining the crate's motion. The worker initially assumes no friction, calculating an initial speed of 5 m/s, but the crate only travels 1.6 meters up the ramp before sliding back down. The key conclusions are that the work done by friction is negative for both the uphill and downhill movements, and that the work done by a force is path-dependent, contradicting the notion that work is zero if the initial and final positions are the same.

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  • Understanding of Newton's laws of motion
  • Familiarity with the concepts of work and energy
  • Knowledge of frictional forces and their effects on motion
  • Basic calculus for evaluating integrals and understanding path dependence
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  • Study the principles of work and energy in mechanical systems
  • Learn about the effects of friction in different scenarios, including inclined planes
  • Explore the concept of conservative vs. nonconservative forces in physics
  • Investigate the mathematical formulation of work done by a force using integrals
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Students and educators in physics, mechanical engineers, and anyone interested in understanding the dynamics of forces and motion, particularly in the context of friction and energy conservation.

henry3369
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We want to slide a 12-kg crate up a 2.5-m-long ramp inclined at . A worker, ignoring friction, calculates that he can do this by giving it an initial speed of 5 m/s at the bottom and letting it go. But friction is not negligible; the crate slides only 1.6m up the ramp, stops, and slides back down (Fig. 7.11a). (a) Find the magnitude of the friction force acting on the crate, assuming that it is constant. (b) How fast is the crate moving when it reaches the bottom of the ramp?

So I'm mainly confused about part b. I know how to get the answer, but I'm confused about why I include the distance traveled up the ramp as well as down the ramp when finding the work done by friction. If I set the starting position at to be the top of the ramp to solve the problem, the block only travels a distance of 1.6 meters not 1.6*2 meters so why do I include the entire path traveled? Also wouldn't work done by friction be zero if you do include the entire distance traveled because Work = Force * displacement and since the stating position is the same as the ending position, the displacement is zero?
 
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Oh. I read the solution incorrectly. But I'm still curious if work done by friction is zero if the initial and final position is the same.
 
The direction of the friction force is opposite to the direction of motion for both the uphill slide and the downhill return. That means that the work done by friction on the crate will be negative for both halves of the round trip.
 
The statement that work is force*displacement is not generally true as you have perhaps just discovered. Generally speaking, the work done by a force depends on the path it takes between two points; the total work done by a force is the integral of the dot product of the force and its differential displacement evaluated along the path.

Work done by conservative forces is independent of the path (this is what defines a conservative force). All conservative forces can be derived from potential functions which is why conservative forces have corresponding potential energies. The work done by friction is certainly not path independent; if you go around in a circular path with zero displacement the force of friction will still have done work. Friction is a nonconservative force.

Interestingly (at least as far as I know), all forces are fundamentally conservative - including friction. The work done by friction ends up as kinetic energy of the atoms on the surfaces in contact. Obviously, it would be impossible (or at least insuperably difficult) to account for all the molecular motions happening with friction. These hidden degrees of freedom are accounted for by introducing the concept of nonconservative forces.
 

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