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A block of mass 2 kg moving at 10 m/s slides onto the left end of a…
Question
A block of mass 2 kg moving at 10 m/s slides onto the left end of a long plank of mass 8 kg that rests on a frictionless horizontal floor. The coefficient of friction between block and plank is 0.5; the floor is frictionless. Taking g = 10 m/s^2, find the total distance (in metres) the block slides relative to the plank before they move together with a common velocity.
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Step-by-step solution
Friction force f = mu m g = 0.5(2)(10) = 10 N.
Block decelerates at a_b = -f/m = -5 m/s^2; plank accelerates at a_p = f/M = 10/8 = 1.25 m/s^2.
They reach common velocity when 10 - 5t = 1.25t, giving t = 10/6.25 = 1.6 s.
Relative initial speed = 10 m/s, relative deceleration = 5 + 1.25 = 6.25 m/s^2.
Relative sliding stops when relative velocity = 0, after relative distance = u^2/(2 a_rel) = 100/(2*6.25) = 8 m.
Final answer8
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