Posted by COFFEE on Monday, February 26, 2007 at 4:11pm.
A stationary block explodes into two pieces L and R that slide across a frictionless floor and then into regions with friction, where they stop. Piece L, with a mass of 2.0 kg, encounters a coefficient of kinetic friction µL = 0.40 and slides to a stop in distance dL = 0.15 m. Piece R encounters a coefficient of kinetic friction µR = 0.50 and slides to a stop in distance dR = 0.20 m. What was the mass of the original block?
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* Physics - bobpursley, Sunday, February 25, 2007 at 5:24pm
the momentum of L and R are equal.
energyL= mu*mg*distance
then solve for velocity L (KE=energy)
knowing velocityL, you can use momentum to find veloicty R.
* Physics - COFFEE, Sunday, February 25, 2007 at 11:29pm
I did:
energyL = mu*mg*distance
energyL = (.4)(2)(9.8)(.15)
energyL = 1.176
Then,
KE = 1/2mv^2
1.176 = 1/2(2)v^2
v = 1.08 m/s
Then which equation do I use???
So the momentum sum is zero.
1.08*2+Vr*massR=0
You have to solve for Vr the same way.
1/2 Mr*Vr^2= Mur*Mr*g*Dr notice the Mr divide out, so solve for Vr. Put that in the equation
1.08*2+Vr*massR=0
solve for mass r.
Then total mass is massR+2
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