Two Blocks on two inclines(Friction)

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The discussion centers on a physics problem involving two blocks on ramps, one frictionless and the other with friction, both inclined at 30 degrees. The block on the frictionless ramp travels further than the block on the friction ramp, specifically 0.625 times the distance. The student is attempting to find the coefficient of sliding friction for the second ramp using the equations of motion and acceleration. The accelerations for both ramps are correctly identified, but there was a mistake in the equation setup regarding parentheses. The corrected equation allows for solving the problem accurately.
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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) .

Find the coefficient of sliding friction between the block and ramp 2.

[PLAIN]https://online-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys100/fall10/homework/07/IE_friction_fraction/frictionfraction.gif



Homework Equations


Ff=muFn
Vf^2=Vi^2+2ad



The Attempt at a Solution



I know that
Vf^2=Vi^2+2ad

and my two formulas for acceleration are
a1 = -g*sin(30)
a2 = -g*sin(30) - μg*cos(30)

since Vi=0
Vf^2=2ad
since the Vf will be the same, the equations can be set equal to each other

2(-g)sin30(d)=2(-g)sin(30)-μgcos(30)(.625d)

I have tried this problem so many times and arrive at a final answer involving two variables. any help for finishing the problem would be greatly appreciated.
 
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Are my formulas correct at least?
 
ryan1180 said:
and my two formulas for acceleration are
a1 = -g*sin(30)
a2 = -g*sin(30) - μg*cos(30)

The accelerations are correct.

ryan1180 said:
since Vi=0
Vf^2=2ad

The final velocities are zero. The initial velocities are equal.

ryan1180 said:
the equations can be set equal to each other

2(-g)sin30(d)=2(-g)sin(30)-μgcos(30)(.625d)

You forgot the parentheses at the right-hand side around the acceleration. The correct equation is

2(-g)sin30(d)=2((-g)sin(30)-μgcos(30))(.625d)

ehild
 
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