In the figure here, a stationary block explodes into two pieces L and R that sli

Q: In the figure here, a stationary block explodes into two pieces L and R that sli

piece L 0.5*mL*Vl^2 - uL*mL*g*dL = 0 0.5*3.5*vl^2 - 0.46*3.5*9.8*0.46 = 0 vl = 2.03 m/s ---- piece R 0.5*mR*VR^2 - uL*mR*g*dR = 0 0.5*vl^2 - 0.55*9.8*0.46 = 0 vl = 2.23 m/s mL*vL = mR*vR mR = 3.186 kg M = ml + mR = 3.5+3.186 = 6.686 kg

ID: 1285090 • Letter: I

Question

In the figure here, 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 3.5 kg, encounters a coefficient of kinetic friction ?L = 0.46 and slides to a stop in distance dL = 0.46 m. Piece R encounters a coefficient of kinetic friction ?R = 0.55 and slides to a stop in distance dR = 0.46 m. What was the mass of the block?

see the image http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c09/qu_09_C.gif

Explanation / Answer

piece L

0.5*mL*Vl^2 - uL*mL*g*dL = 0

0.5*3.5*vl^2 - 0.46*3.5*9.8*0.46 = 0

vl = 2.03 m/s

----

piece R

0.5*mR*VR^2 - uL*mR*g*dR = 0

0.5*vl^2 - 0.55*9.8*0.46 = 0

vl = 2.23 m/s

mL*vL = mR*vR

mR = 3.186 kg

M = ml + mR = 3.5+3.186 = 6.686 kg

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