Initial and Final Energy problem

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SUMMARY

The discussion focuses on a physics problem involving a 2.4 kg piece of wood sliding on a surface with kinetic friction. The wood starts from rest at a height of 4.0 m and encounters a rough horizontal surface with a friction coefficient of 0.27. The key equations used include the kinetic energy formula (E_k = (1/2)mv^2), the work done by friction (E_f = mg(u_k)d), and gravitational potential energy (E_g = mgh). The correct approach involves equating initial gravitational energy to the work done by friction to determine the distance the wood travels before coming to rest.

PREREQUISITES
  • Understanding of gravitational potential energy (E_g = mgh)
  • Knowledge of kinetic energy (E_k = (1/2)mv^2)
  • Familiarity with the work-energy theorem
  • Concept of kinetic friction and its coefficient
NEXT STEPS
  • Calculate the distance traveled by the wood using the equation mgh = Ugdm
  • Explore the work-energy theorem in the context of frictional forces
  • Review the principles of energy conservation in mechanical systems
  • Investigate the effects of varying friction coefficients on motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy concepts, as well as educators looking for examples of energy conservation and friction in real-world applications.

YamiBustamante
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Homework Statement


[/B]
A 2.4 kg piece of wood slides on the surface shown in the figure . The curved sides are perfectly smooth, but the rough horizontal bottom is 31m long and has a kinetic friction coefficient of 0.27 with the wood. The piece of wood starts from rest 4.0m above the rough bottom.

a) Where will this wood eventually come to rest?

b) For the motion from the initial release until the piece of wood comes to rest, what is the total amount of work done by friction?

Homework Equations


E_k = (1/2)mv^2
E_f = mg(u_k)d
E_g = mgh

The Attempt at a Solution



I think I have an idea for part a but I'm not sure. Would this be correct?

Initial Energy = Final Energy
Potential Gravitational Energy = Friction Energy
mgh = Ugdm
(2.4kg)(9.8m/s^2)(4.0m) = 0.27(9.8m/s^2)(2.4kg)d
And then I figured that I would solved for d...
Would that be correct? And for part b would I use the work kinetic energy theorem even though it has friction?
 

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I think you got it.
 

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