A small block with a mass of 0.140 kg is attached to acord passing through a hol

Question

A small block with a mass of 0.140 kg is attached to acord passing through a hole in a frictionless, horizontal surface(the figure ). The block is originally revolving at a distance of0.50 m from the hole with a speed of 0.20 m/s. The cordis then pulled from below, shortening the radius of the circle inwhich the block revolves to 0.15 m. At this new distance, thespeed of the block is observed to be 0.67 m/s. How much work was done by the person who pulled thecord? I already found T2 = 0.41 N and T1= .011N A small block with a mass of 0.140 kg is attached to acord passing through a hole in a frictionless, horizontal surface(the figure ). The block is originally revolving at a distance of0.50 m from the hole with a speed of 0.20 m/s. The cordis then pulled from below, shortening the radius of the circle inwhich the block revolves to 0.15 m. At this new distance, thespeed of the block is observed to be 0.67 m/s. How much work was done by the person who pulled thecord? I already found T2 = 0.41 N and T1= .011N

Explanation / Answer

The work is done against the centripetal force Fc Fc= m V2/R Since V is depenmdent on R we need the velocity with angularvelocity V= R we have then Fc= m (R)2/R Fc= m ()2R or Fc avg= integral (m ()2 dR) We can integrate this from R2= 0.5 to R1=0.15 and wehave Fcavg = (m ()2 ( R2 -R1) since at R= 0.15 m 0.67 m/s then =V/R =0.67/0.15=4.47 rad/s Fcavg = (0.140(4.47)2 ( 0.50 -0.15) Fcavg = 0.98 N W=F d then W= integral ( (m ()2 RdR) W=0.5(m ()2 ( R22 -R12) W= 0.5(0.140(4.47)2 ( 0.502-0.152) W=0.32 J

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