<p>We wrap a light, flexible cable around a solid cylinder with mass 10.0 kg &#1

Question

<p>We wrap a light, flexible cable around a solid cylinder with mass 10.0 kg &#160;and diameter 40.0 cm. We tie the free end of the cable to a block of mass 11.0 kg. The 11.0 kg-mass is released from rest and falls, causing the cylinder to turn about a frictionless axle through its center. As the block falls, the cable unwinds without stretching or slipping, turning the cylinder.<br /><br />How far will the mass have to descend to give the cylinder 260 J of kinetic energy?</p>

Explanation / Answer

Given Mass of solid cylinder , m = 10kg Radius of cylinder , r = 0.2 m Mass of block , M = 11 kg Kinetic energy , KE = 260 J Net force on the block is, F = Mg - T ---1 Net torque acting is = I r * T = I r *T = (1/2m r^2) (a /r) T = ma /2 substitute T in eq 1 Ma= Mg - ma /2 a = 11kg *9.8 m/s^2 / (11kg + (10kg/2)) a = 6.73 m/s^2 Net force on the block is, F = Ma = 11kg *6.73 m/s^2 F = 74.03 N ------------------------------------------------------------ By work energy theorem F *s = KEf - KE i 74.03N *s = 260 J s = 3.5 m is the distance the mass descended

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