A small block with a mass of 0.0900 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Fig). The block is originally revolving at a distance of 0.40 m from the hole with a speed of 0.70 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? = 0.70 m/s? (b) What is the tension in the cord in the final situation when the block has speed ?v? = 2.80 m/s? (c) How much work was done by the person who pulled on the cord? Figure:

Solution 75P Step 1: Data given Mass of the block m = 0.090 kg Radius of circle 1 r = 0.40 m 1 Velocity along circular path v =10.70 m/s Radius of circle 2 r 2 0.10 m Velocity along circular path v =22.80 m/s Let us find the acceleration of the block around the circular path It is given by a = v /r For circle 1 We have 2 a1= (0.70 m/s) /0.40 m 2 a1= 0.0122 m/ s Similarly acceleration for circle 2 a2= (2.80 m/s) /0.10 m a = 78.4 m/s 2 2 Step 2 : Tension for string before pulling is T 1 mass × a 1 Substituting we get T 1 0.090 kg × 0.0122 m/ s 2 T 1 1.098 × 10 3 This can written as 0.11 N Thus we have tension before pulling as0.11 N Tension after pulling is T 2 mass × a 2 Substituting we get T = 0.090 kg × 78.4 m/s 2 2 T = 7.056 2 This can be written as 7.1 N Thus we get tension after pulling the string as 7.1 N