Coefficient of friction of a ramp

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SUMMARY

The discussion centers on calculating the coefficient of friction for two ramps at a 30-degree angle, where Ramp 1 is frictionless and Ramp 2 has friction. The block on Ramp 2 travels 0.625 times the distance of the block on Ramp 1 before stopping. The equations used include Ff = μFn and v² = v₁² + 2ad, leading to the derivation of the coefficient of friction (μ) based on the relationship between the distances traveled and the forces acting on the blocks.

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  • Understanding of Newton's laws of motion
  • Familiarity with the concepts of friction and normal force
  • Knowledge of kinematic equations
  • Basic trigonometry, particularly sine and cosine functions
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maladroit
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Homework Statement



A student has two ramps, both of which are at an angle of 30o. Ramp 1 is frictionless and ramp 2 has friction. The student also has two blocks, one for each ramp. She pushes the blocks up the ramps with the same initial velocity. The block on ramp 2 only travels a fraction f = 0.625 as far as the block on ramp 1 before coming to a stop (i.e. d2 = 0.625*d1) .

Homework Equations



Ff=\muFn
v2^2-v1^2=2ad


The Attempt at a Solution



I am way stuck on this!
 
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Hi maladroit, welcome to PF.
While posting any problem, state it completely as it is given in the home work.
In the above problem what is required? Are the two blocks identical?
 
That is exactly as the problem is written, the only thing I forgot to include was that we are looking for the coefficient of friction. Sorry about that!
 
In the ramp 1, the retardation is g*sinθ.
Initial velocity is v and the final velocity is zero. So
v^2 = 2*g*sinθ*d
In the second case, the normal reaction is mgcosθ and frictional force = μmgcosθ.
So the net retardation is (g*sinθ + μgcosθ)
Now write down the expression for v^2 and solve for μ.
 
Thank you so much for the help!
 

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