A cord is used to vertically lower an initially stationary block of mass M = 45

Question

A cord is used to vertically lower an initially stationary block of mass M = 45 kg at a constant downward acceleration of g/3. When the block has fallen a distance d = 6.0 m, find the work done by the cord's force on the block. AND Find the work done by the gravitational force on the block. AND Find the kinetic energy of the block. AND Find the speed of the block. So four parts please, 1. The work done by the cords force on the block, 2. work done by gravitational force on the block, 3. Kinetic energy of the block, and 4. Speed of the block. Thanks!

Explanation / Answer

Let T be the force of the cord on the block so m*g - T = m*a = m*g/3 So T = m*(g -g/3) = 40*(9.80 - 9.80/3) = 261.3N W (cord) = -261.3N*6.0m = -1568 J (It is negative because the direction of the force is opposite to the motion) 2. W = m*g*d = 40*9.8*6 = 2352 J 3. From work- energy W = ?K so K = 2352 - 1568 = 784J 4. K = 1/2*m*v^2 so v = sqrt(2*K/m) = sqrt(2*784/40) = 6.26m/s

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