A small block with a mass of 0.160 is attached to a cord passing through a hole

Question

A small block with a mass of 0.160 is attached to a cord passing through a hole in a frictionless, horizontal surface (the figure ). The block is originally revolving at a distance of 0.50 from the hole with a speed of 0.20 . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.10 . At this new distance, the speed of the block is observed to be 1.00 .

A)What is the tension in the cord in the original situation when the block has speed = 0.20 ?
Express your answer using two significant figures.

B)What is the tension in the cord in the final situation when the block has speed = 1.00 ?
Express your answer using two significant figures.

C)How much work was done by the person who pulled on the cord?
Express your answer using two significant figures.

Explanation / Answer

Given Mass of the block is , m = 0.160 kg Initial radius , r1 = 0.5 m Speed of the block at a distance r1 is , v1 = 0.2 m/s Final radius , r2 = 0.1 m Speed of the block at a distance r2 is , v2 = 1 m/s A) The tension in the string when v1 = 0.2 m/s is        T1 = mv12 /r1 = (0.160 kg ) (0.2 m/s)2/0.5 m         T1 = 0.013 N B) The tension in the string when v2 = 1.0 m/s is        T2 = mv22 /r2 = (0.160 kg ) (1.0 m/s)2/0.1 m        T2 = 1.6 N C) Work done by the person is       W = KE1 - KE2              = 1/2 m (v22-v12) = 0.5 *0.160 kg * [(1.0 m/s)^2-(0.2 m/s)^2]       W = 0.0768 J Speed of the block at a distance r2 is , v2 = 1 m/s A) The tension in the string when v1 = 0.2 m/s is        T1 = mv12 /r1 = (0.160 kg ) (0.2 m/s)2/0.5 m         T1 = 0.013 N B) The tension in the string when v2 = 1.0 m/s is        T2 = mv22 /r2 = (0.160 kg ) (1.0 m/s)2/0.1 m        T2 = 1.6 N C) Work done by the person is       W = KE1 - KE2              = 1/2 m (v22-v12) = 0.5 *0.160 kg * [(1.0 m/s)^2-(0.2 m/s)^2]       W = 0.0768 J              = 1/2 m (v22-v12) = 0.5 *0.160 kg * [(1.0 m/s)^2-(0.2 m/s)^2]       W = 0.0768 J

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