The puck in the figure below has a mass of 0.126 kg. Its original distance from

Question

The puck in the figure below has a mass of 0.126 kg. Its original distance from the center of rotation is 40.6 cm, and the puck is moving with a speed of 80.7 cm/s. The string is pulled downward 16.7 cm through the hole in the frictionless table. Determine the work done on the puck.

Explanation / Answer

THe angular vel of the puck will be same, but rad of rotn decreases, so velocity decreases => Kinetic energy of it decreases. WORK DONE BY THE FORCE = CHANGE IN KE v1/r1=v2/r2 (SINCE ang vel w = v1/r1=v2/r2) 80.7/40.6 = v2 /(40.6-16.7) => v2 = 47.51 cm/s SO NOW ANSWER = CHANGE IN KE = 1/2 * m * (v2^2 - v1^2) = -0.0268 J (COnver cm to metres) Hope you like my clear and detailed explanation :) (Im sorry if i faulted with any calculations, no calculator :( Had to do them by hand)

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