Block slides along an elevated track

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SUMMARY

The discussion focuses on a physics problem involving a block sliding along an elevated track with a flat section of length L = 0.2m and an initial height h = 0.1m. The block experiences gravitational potential energy calculated as U = mgh = m * 9.8 * 0.1 = 0.98m. The track's flat portion has a friction coefficient of μk = 0.15, and the curved section is frictionless. The solution requires applying the concept of pseudowork, defined as ΔKtrans = (1/2)mv² = ∫ F·dr, where the frictional force is used to determine the distance the block travels before coming to rest.

PREREQUISITES
  • Understanding of gravitational potential energy (U = mgh)
  • Familiarity with the concept of kinetic energy (K = (1/2)mv²)
  • Knowledge of friction coefficients and their application in physics problems
  • Basic understanding of work-energy principles, including true work and pseudowork
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  • Study the application of pseudowork in mechanics problems
  • Learn about the effects of friction on motion in physics
  • Explore energy conservation principles in systems with friction
  • Practice solving problems involving kinetic and potential energy transformations
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Students studying physics, particularly those focusing on mechanics and energy concepts, as well as educators seeking to enhance their teaching of work-energy principles.

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Homework Statement


A block slides along an elevated track. The flat part has a length L = 0.2m and the object is released from rest from height h = 0.1m. The curved portion of the track is frictionless, but the flat part is rough with a friction coefficient μk = 0.15. Where does the object finally come
to rest? *Hint: use pseudowork to account for friction.
http://img694.imageshack.us/img694/9565/dsfm.jpg


Homework Equations


U = mgh =m9.8*0.1


The Attempt at a Solution


i know that the object starts with U = mgh =m9.8*0.1=m0.98 but i have no idea what to do next.. and what's pseudowork?
 
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I had to google it myself, but apparently pseudowork is given by

[tex] \Delta K_{trans}=\frac{1}{2}mv_{cm}^2=\int \mathbf{F}\cdot d\mathbf{r}[/tex]

where the cm=center of mass. True work is given by

[tex] W=\int\mathbf{F}\cdot d\mathbf{r}[/tex]

My guess is you'll be using the frictional force for [itex]\mathbf{F}=\mu_k\mathbf{F}_{normal}[/itex] and solving for [itex]\mathbf{r}[/itex] since you will know everything else.
 

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