Inclined plane, Calculate friction force

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SUMMARY

The discussion focuses on calculating the friction force on an inclined plane using the work-energy theorem. The user initially calculated the friction force as 12.28N but later corrected it to 11N after providing detailed calculations. The equations utilized include Wnet = Ek and the breakdown of work done against gravity and friction. The final confirmation from another participant validates the corrected friction force calculation.

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  • Understanding of the work-energy theorem
  • Knowledge of inclined plane physics
  • Familiarity with basic trigonometry (e.g., sine function)
  • Ability to perform algebraic manipulations of equations
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  • Study the work-energy theorem in detail
  • Learn about forces on inclined planes and their calculations
  • Explore advanced friction concepts in physics
  • Practice solving problems involving inclined planes and friction forces
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for practical examples of applying the work-energy theorem in real-world scenarios.

TheRedDevil18
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Homework Statement



work.jpg


Homework Equations



Wnet = Ek

The Attempt at a Solution



1) That was fine.
2) I just want to confirm if this answer is right?, I got 12.28N. Used work energy theorem.

Sorry for double post, edited the title
 
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TheRedDevil18 said:

Homework Statement



work.jpg


Homework Equations



Wnet = Ek

The Attempt at a Solution



1) That was fine.
2) I just want to confirm if this answer is right?, I got 12.28N. Used work energy theorem.

Sorry for double post, edited the title

Seems a bit high. Pls post your working.
 
haruspex said:
Seems a bit high. Pls post your working.

Sorry my mistake, forgot something in my calculation. Anyway here's my working

Wnet = Ek
Wg + Wf + Wn = 1/2*m(vf^2 - vi^2)
(5*9.8*sin30*20) - 20Ff + 0 = 1/2*5(12^2 - 6^2)...That 20 is the length of the incline
Ff = 11N...Now correct?
 
TheRedDevil18 said:
Sorry my mistake, forgot something in my calculation. Anyway here's my working

Wnet = Ek
Wg + Wf + Wn = 1/2*m(vf^2 - vi^2)
(5*9.8*sin30*20) - 20Ff + 0 = 1/2*5(12^2 - 6^2)...That 20 is the length of the incline
Ff = 11N...Now correct?

Yes, that looks right.
 
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