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CP A 0.300-kg potato is tied to a string with length 2.50

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 59P Chapter 7

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 59P

CP A 0.300-kg potato is tied to a string with length 2.50 m, and the other end of the string is tied to a rigid support. The potato is held straight out horizontally from the point of support, with the string pulled taut, and is then released. (a) What is the speed of the potato at the lowest point of its motion? (b) What is the tension in the string at this point?

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Step 1 of 3

Solution 59P To solve this question, we shall have to apply the law of conservation of energy. If the potato is at the lowest point of motion, its kinetic energy and potential energy at this position are equal. 1mv = mgh…..(1), where m is the mass of the potato, g is acceleration due to gravity, 2 v is the speed of the potato at the lowest point h is the height moved by the potato. In this case, the height moved is nothing but the length of the string. This can be understood from the figure given below. Given that, mass m = 0.300 kg Length of the string h = 2.50 m Acceleration due to gravity g = 9.80 m/s 2 (a) Therefore, from equation (1), v = 2gh 2 2 v = 2 × 9.80 m/s × 2.50 m 2 2 2 v = 49 m /s v = 7.00 m/s Therefore, the speed of the potato is 7.00 m/s when it reaches the lowest point of motion. (b) The tension in the string at this point is given by the expression, 2 2 T = mg + mv , (here mv is the centripetal force term for circular motion) h h 2 0.300 kg×(7.00) m /s T = 0.300 kg × 9.8 m/s + 2.50 m T = 8.82 N The tension in the string is 8.82 N.

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Chapter 7, Problem 59P is Solved
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Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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CP A 0.300-kg potato is tied to a string with length 2.50