A small block with a mass of m = 0.120 kg is attached to a cord passing through a hole in a frictionless, horizontal surface. 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 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 a speed of: v = 0.70 m/s?
b) What is the tension in the cord in the final situation when the block has speed of:
v = 2.80 m/s?
c) How much work was done by the person who pulled the cord?
2 answers
did u get the answer?????
a) (m*v^2)/d=(0.12*0.7^2)/0.4=0.147 N
b)(m*v^2)/d=(0.12*2.80^2)/0.1=9.408 N
c) W=deltaKE=1/2mv^2 -1/2mu^2 =1/2 x 0.12 x [(2.8)^2 - (0.70)^2] =0.44 J
b)(m*v^2)/d=(0.12*2.80^2)/0.1=9.408 N
c) W=deltaKE=1/2mv^2 -1/2mu^2 =1/2 x 0.12 x [(2.8)^2 - (0.70)^2] =0.44 J
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