Use the work-energy theorem to solve each of these

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

Problem 19E Chapter 6

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 19E

Use the work-energy theorem to solve each of these problems. You can use Newton’s laws to check your answers. Neglect air resistance in all cases (a) A branch falls from the top of a 95.0-m-tall redwood tree, starting from rest. How fast is it moving when it reaches the ground? (b) A volcano ejects a boulder directly upward 525 m into the air. How fast was the boulder moving just as it left the volcano? (c) A skier moving at 5.00 m/s encounters a long, rough horizontal patch of snow having coefficient of kinetic friction 0.220 with her skis. How far does she travel on this patch before stopping? (d) Suppose the rough patch in part (c) was only 2.90 m long? How fast would the skier be moving when she reached the end of the patch? (e) At the base of a friclionless icy hill that rises at 25.0° above the horizontal, a toboggan has a speed of 12.0 m/s toward the hill. How high vertically above the base will it go before stopping?

Step-by-Step Solution:
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Solution 19E The work-energy theorem states that the change in kinetic energy of a particle is equal to the work done by the net force on the particle. Mathematically, Work done = change in kinetic energy or W = K …..(1) (a) iven that the branch falls from the top of a 95.0 m tall tree from rest. So, h = 95.0 m 1 2 1 2 Therefore, its change in kinetic energy K = mv 2mu , wh2re m is the mass of the branch, u is its initial speed and v is the final speed or the speed when it reaches the ground. Given that, u = 0 Therefore, K = mv …..(2) 2 Now, work done by gravitational force W = mgh, where g is the acceleration due to gravity and h is the height. Now, equating W and K , we get 1 2 2mv = mgh v = 2gh v = 2 × 9.8 × 95.0 v = 43.2 m/s When the branch reaches the ground, its speed will be 43.2 m/s. (b) The height reached by the boulder is h = 525 m At the maximum height, the speed of the boulder is v = 0 We have to calculate the initial speed u of the boulder. From the work-energy theorem, 2m(u v ) = mgh, m is the mass of the boulder u v = 2gh 2 u = 2gh , since v = 0 u = 2h u = 2 × 9.8 × 525 m/s u = 10290 m/s u = 101 m/s So,...

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Chapter 6, Problem 19E is Solved
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Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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Use the work-energy theorem to solve each of these

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