A block of massm=2.00kg rests on the left edge of a block of length L=3.00 m and

Question

A block of massm=2.00kg rests on the left edge of a block of length L=3.00 m andmass M=8.00kg. The coefficent of the kinetic friction between thetwo blocks is k=0.300, and the surface on which the 8.00kgblock rests is frictionless. A constant horizontal force ofmagnitude F=12.0N is applied to the 2.00kg block, setting it inmotion. Howlong will it take before this block makes it to the right side ofthe 8.00kg block?(Hint: Both blocks are set in motion when theforce vector F is applied.) How fardoes the 8.00kg block move in the process? A block of massm=2.00kg rests on the left edge of a block of length L=3.00 m andmass M=8.00kg. The coefficent of the kinetic friction between thetwo blocks is k=0.300, and the surface on which the 8.00kgblock rests is frictionless. A constant horizontal force ofmagnitude F=12.0N is applied to the 2.00kg block, setting it inmotion. Howlong will it take before this block makes it to the right side ofthe 8.00kg block?(Hint: Both blocks are set in motion when theforce vector F is applied.) How fardoes the 8.00kg block move in the process?

Explanation / Answer

Fm = 12 N     force onsmall block
FM = m g = .3 * 2 * 9.8 = 5.88 N   = Ff   force on large blockdue to friction Fm - Ff = 6.12 N   = mam   net force on smallblock     am = 6.12 / 2 = 3.06 m /s2 Ff = M a M    aM = 5.88 / 8 = .735 m / s2 1/2 am t2 = 1/2 aMt2 + s     where is the extradistance that block m travels in time t t2 = 2 s / (am -aM)    time for block m to travel extradistance s The problem isn't worded too explicitly but apparently 3 m isthe distance to use for s In that case t2 = 6 / (3.06 - .735) =2.58    and t = 1.61 sec sM = 1/2 aM t2 = (.735 / 2) *1.612 = .948 m 1/2 am t2 = 1/2 aMt2 + s     where is the extradistance that block m travels in time t t2 = 2 s / (am -aM)    time for block m to travel extradistance s The problem isn't worded too explicitly but apparently 3 m isthe distance to use for s In that case t2 = 6 / (3.06 - .735) =2.58    and t = 1.61 sec sM = 1/2 aM t2 = (.735 / 2) *1.612 = .948 m

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