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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 6 / Problem 75P

Sears and Zemansky's University Physics with Modern Physics | 13th Edition | ISBN: 9780321696861 | Authors: Hugh D. Young; Roger A. Freedman; A. Lewis Ford

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Answer: A small block with a mass of 0.0900 kg is attached

Chapter 6, Problem 75P

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QUESTION:

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. P6.75). 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 \(v = 0.70 \mathrm{\ m}/\mathrm{s}\)? (b) What is the tension in the cord in the final situation when the block has speed \(v = 2.80 \mathrm{\ m}/\mathrm{s}\)? (c) How much work was done by the person who pulled on the cord?

Text Transcription:

v = 0.70 m/s

v = 2.80 m/s

Questions & Answers

QUESTION:

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. P6.75). 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 \(v = 0.70 \mathrm{\ m}/\mathrm{s}\)? (b) What is the tension in the cord in the final situation when the block has speed \(v = 2.80 \mathrm{\ m}/\mathrm{s}\)? (c) How much work was done by the person who pulled on the cord?

Text Transcription:

v = 0.70 m/s

v = 2.80 m/s

ANSWER:

Solution 75P

Step 1:

Data given

Mass of the block $m\ =\ 0.090\ kg\$

Radius of circle 1 ${r}_{1}\ =\ 0.40\ m\$

Velocity along circular path ${v}_{1}=\ 0.70\ m/s$

Radius of circle 2 ${r}_{2}\ =\ 0.10\ m\$

Velocity along circular path ${v}_{2}=\ 2.80\ m/s$

Let us find the acceleration of the block around the circular path

It is given by

$a=\ {v}^{2}/{r}_{\ }$

For circle 1

We have

${a}_{1}=\ {(0.70\ m/s)}^{2}/{0.40\ m\ }_{\ }$

${a}_{1}=\ 0.0122\ m{/}^{\ }{s}^{2}$

Similarly acceleration for circle 2

${a}_{2}=\ {(2.80\ m/s)}^{2}/0.10\ m\$

${a}_{2}=\ 78.4\ m/{s}^{2}$

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