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?

For Further Reading

* 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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