Work-Energy: Find the angle of the inclined plane

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SUMMARY

The discussion focuses on calculating the angle of an inclined plane using work-energy principles. A block with a mass of 5 kg slides down an inclined plane with a height of 1.5 m and a coefficient of kinetic friction of 0.35, achieving a final speed of 4.2 m/s. The net work equation is established as {W^{{\rm{net}}}} = mg(\Delta y) - {\mu _k}mgd\cos (\varphi ) = \frac{1}{2}mv_2^2. The calculated angle of the inclined plane is approximately 41.186 degrees, indicating the solution is plausible based on the provided parameters.

PREREQUISITES
  • Understanding of work-energy principles in physics
  • Familiarity with kinetic friction and its coefficient
  • Knowledge of trigonometric functions, specifically sine and cotangent
  • Ability to manipulate equations involving mass, height, and velocity
NEXT STEPS
  • Review the derivation of the work-energy theorem in physics
  • Study the effects of kinetic friction on inclined planes
  • Learn about the law of sines and its applications in physics problems
  • Explore different methods for calculating angles in inclined plane scenarios
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and inclined plane problems, as well as educators seeking to clarify work-energy concepts.

epr2008
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Homework Statement


A block with mass 5 kg slides down and inclined plane with height 1.5 m and coefficient of kinetic friction .35. The final speed of the block is 4.2 m/s. Find the angle of the inclined plane.

Homework Equations


[tex]{W^{{\rm{net}}}} = {W^{{\rm{grav}}}} - {W^{{\rm{fric}}}} = {K_2} - {K_1}[/tex]

The Attempt at a Solution


This has taken me forever and i still don't think it's right but here it goes.

I have

[tex]{W^{{\rm{net}}}} = mg(\Delta y) - {\mu _k}mgd\cos (\varphi ) = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2[/tex]

Then since initial velocity is 0

[tex]{W^{{\rm{net}}}} = mg(\Delta y) - {\mu _k}mgd\cos (\varphi ) = \frac{1}{2}mv_2^2[/tex]

I'm sure there is another way to find d but i couldn't think of any besides the law of sines.

[tex]\frac{d}{{\sin (90)}} = d = \frac{{\Delta y}}{{\sin (\varphi )}}[/tex]

So now I have

[tex]{W^{{\rm{net}}}} = mg(\Delta y) - {\mu _k}mg(\Delta y)\cot (\varphi ) = \frac{1}{2}mv_2^2[/tex]

And solving

[tex]\cot (\varphi ) = \frac{{mg(\Delta y) - \frac{1}{2}mv_2^2}}{{{\mu _k}mg(\Delta y)}}[/tex]

I am using the calculator on my phone so I don't know if it is right but

[tex]\varphi = 41.186^\circ[/tex]

Would anyone care to help me out and tell me if I'm at least in the ballpark?
 
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