Two blocks are attached over a massless, frictionless pulley as shown.?
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Initially the left mass(5kg) is touching the floor, the right(15kg) is 2m up and they are released from rest. a.What is the acceleration of the blocks? What is the tension in the cord? b. Just before the right hits the floor, what is the total kinetic energy of the blocks? What is the speed of the two blocks?
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Answer:
Just draw the free-body diagram and we get, a) T - m1g = m1a m2g - T = m2a solving this we get. a = [(m2 - m1)/(m2 + m1)]g = 10/20*20 = 5 m/s^2 T = 2m1m2g/(m1 + m2) = 2*5*15*10/20 = 75 N b) K.E = decrease in potential energy = (m2-m1)gh = (15-5)*10*2 = 200 J now let each block has speed v.. so, .5*(m1+m2)*v^2 = 200 or v = sqrt(20) = 4.47 m/s hope this helps..
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