A small block with a mass of 0.105 kg is attached to a cord passing through a ho

Question

A small block with a mass of 0.105 kg is attached to a cord passing through a hole in a horizontal frictionless surface as shown in the figure below. The block is originally revolving at a distance of 0.31 m from the hole with a speed of 0.61 m/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.10 m. At this new distance the speed of the block is observed to be 2.80 m/s. (a) What is the tension in the cord in the original situation when the block has speed v = 0.61 m/s? N (b) What is the tension in the cord in the final situation when the block has speed v = 2.80 m/s? N (c) How much work was done by the person who pulled on the cord? J

Explanation / Answer

when the block is revolving, it has a cnetrifugal force and the tension in the thread actring on it, for equilibrium bothmust be equal

a) tesnion T = mv2/r (centrifugal force)

     m = 0.105 kg , r = 0.31, v = 0.61 m/s

      T = 0.126 N

b) r= 0.1m , v = 2.8 m/s

        T = 0.105*2.82/0.1 = 8.232 N

c) Initial KE of the block   = 0.5 mv2 = 0.5 *0.105 *0.612 = 0.0186 J

final KE                                               = 0.5*0.105*2.82 = 0.4116 J

The difference in KE (0.4116 - 0.0186) 0.393 J is the work done

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