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Answer: A small block with a mass of 0.0900 kg is attached
Chapter 6, Problem 75P(choose chapter or problem)
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
Radius of circle 1
Velocity along circular path
Radius of circle 2
Velocity along circular path
Let us find the acceleration of the block around the circular path
It is given by
For circle 1
We have
Similarly acceleration for circle 2