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A mass m slides down a smooth inclined plane from an

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

Solution for problem 22E 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 22E

A mass m slides down a smooth inclined plane from an initial vertical height h, making an angle ? with the horizontal. (a) The work done by a force is the sum of the work done by the components of the force. Consider the components of gravity parallel and perpendicular to the surface of the plane. Calculate the work done on the mass by each of the components, and use these results to show that the work done by gravity is exactly the same as if the mass had fallen straight down through the air from a height h. (b) Use the work–energy theorem to prove that the speed of the mass at the bottom of the incline is the same as if the mass had been dropped from height h, independent of the angle ? of the incline. Explain how this speed can be independent of the slope angle. (c) Use the results of part (b) to find the speed of a rock that slides down an icy frictionless hill, starting from rest 15.0 m above the bottom.

Step-by-Step Solution:
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Solution 22E Here in the diagram below, the innards of the problem is shown. a)The gravitational force is acting downwards and there is no horizontal component of it. The only component is downwards which is the force itself. It’s a well established diagonalised force vector. So, the work done is, W = F.d = Fd cos (90 ) = Fd sin . But by the laws of trigonometry, we can deduce the relationship between d and h. sin = h/d d = h / sin So, the work done can be rewritten as, W = Fd sin = F × h × sin = Fh = mgh. sin Now from work energy theorem we know that, work done = K.E K.Ef i work done = 1/2 mv 0 = 1/2 mv 2 But from the conservation of mechanical energy we also know that, K.E +iP.E = KiE + K.E f f mgh + 0 = 1/2 mv + o 2 2 mgh = 1/2 mv . So, this is proved that, the mass would do the same amount of work which it would have done if falling from a height of “h” freely under gravity. The horizontal components are zero, hence the component...

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Chapter 6, Problem 22E is Solved
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Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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