Posted by Ted on Sunday, October 23, 2011 at 8:56pm.
down a frictionless slide: mg*SinTheta=ma
down a friction slide: mgSinTheta-mu*mg*CosTheta=ma
so if the time for friction is 2x, then its acceleration is half of the original a.
mgSinTheta-mu*mg*CosTheta=1/2 *(sinTheta)mg
solve for mu.
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