A small block with a mass of 0.09kg is attached to a cord passing through a hole
ID: 3898245 • Letter: A
Question
A small block with a mass of 0.09kg is attached to a cord passing through a hole in a frictionless, horizontal surface (As shown in drawing). The block is originally revolving at a distance of 0.4m from the hole with a speed of 0.7m/s. The cord is then pulled from below, shortening the radius of revolution to 0.1m. At this new distance the speed is 2.8m/s. What is the tension in the cord before the cord is pulled? What is the tension in the cord after is pulled (Speed = 2.8 m/s)? How much work was done by the person who pulled on the cord?
Explanation / Answer
BEFORE PULLING
force balancing
T=MV*V/R
= (0.09)*(0.7)*(0.7)/(0.4)
=0.11025N
AFTER PULING
T=MV*V/R
=(0.09)*(2.8)*(2.8)/(0.1)
=7.056N
WORK DONE = INCREASE IN KINETIC ENERGY (WORK BY NON CONSERVATIVE FRICTION FORCE = 0)
=1/2*(0.09)* ( 2.8*2.8 - 0.7*0.7 )
=0.33075 JOULE
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