gain in kinetic energy = (1/2) I omega^2 + (1/2) m v^2
omega r = v so omega = v/r
gain of Ke = (1/2) (.0061) (1.129/.06)^2 + (1/2) .74(1.129)^2
loss of potential energy = .74 (9.81)(.69)
if there were no friction the loss in potential energy = gain in kinetic energy.
if the Ke is less, the difference was burned off by friction.
A block of mass 0.74 kg is suspended by a string which is wrapped so that it is at a radius of 0.060 m from the center of a pulley. The moment of inertia of the pulley is 0.0061 kg·m2. There is friction as the pulley turns. The block starts from rest, and its speed after it has traveled downwards a distance of D= 0.69 m, is 1.129 m/s. Calculate the amount of energy dissipated up to that point.
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