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A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Figure 1). The block is originally revolving at a distance of 0.43 m from the hole with a speed of 0.75 m/s . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.19 m . At this new distance, the speed of the block is 1.70 m/s .What is the tension in the cord in the original situation when the block has speed v0 = 0.75 m/s ?What is the tension in the cord in the final situation when the block has speed v1 = 1.70 m/s ?How much work was done by the person who pulled on the cord?

Question

A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Figure 1). The block is originally revolving at a distance of 0.43 m from the hole with a speed of 0.75 m/s . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.19 m . At this new distance, the speed of the block is 1.70 m/s .

What is the tension in the cord in the original situation when the block has speed v0 = 0.75 m/s ?
What is the tension in the cord in the final situation when the block has speed v1 = 1.70 m/s ?
How much work was done by the person who pulled on the cord?
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Step 1

The tension in the cord when the object is revolving in a circle of radius r with a speed v is equal to the centripetal force and is expressed as,

T Fcentripetal
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T Fcentripetal

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Step 2

The tension in the cord if the speed of the block is 0.75m/s can be calculated using equation 1.

(0.0600kg) (0.75m/s)
(0.43 m)
T =
7.85 X102N
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(0.0600kg) (0.75m/s) (0.43 m) T = 7.85 X102N

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Step 3

The tension in the cord if the speed of the block is...

(0.0600 kg)(1.70m/s)
Т-
(0.19m)
0.91N
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(0.0600 kg)(1.70m/s) Т- (0.19m) 0.91N

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