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Prevention of Hip Injuries. People (especially the

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

Solution for problem 62P Chapter 5

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 62P

Prevention of Hip Injuries?. People (especially the elderly) who are prone to falling can wear hip pads to cushion the impact on their hip from a fall. Experiments have shown that if the speed at impact can be reduced to 1.3 m/s or less, the hip will usually not fracture. Let us investigate the worst-case scenario in which a 55-kg person completely loses her footing (such as on icy pavement) and falls a distance of 1.0 m, the distance from her hip to the ground We shall assume that the person’s entire body has the same acceleration, which, in reality, would not quite be true. (a) With what speed does her hip reach the ground? (b) A typical hip pad can reduce the person’s speed to 1.3 m/s over a distance of 2.0 cm. Find the acceleration (assumed to be constant) of this person’s hip while she is slowing down and the force the pad exerts on it. (c) The force in part (b) is very large. To see whether it is likely to cause injury, calculate how long it lasts.

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Solution 62 2 (a) Acceleration during falling on the ground is g = 9.8 m/s , the distance travelled is h = 1.0 m and the initial velocity is u = 0. So, from dynamics we can write that 2 v = 2gh 2 2 v = 2gh = 2(9.8 m/s )(1.0 m) = 14 m/s Her heap touches the ground with speed 14 m/s . 2 (b) In this case the initial distance speed is u = 14 m/s , final speed is v = 1.3 m/s and the distance travelled is s = 2 cm = 0.02 m Now from the dynamics we know that v = u + 2as 2 2 a = v u2 = (1.3 m) (14 m= 4858 m/s 2 2s 2(0.02 m) Negative sign means the acceleration opposes the velocity. 2 So the acceleration of the person is 4858 m/s . Now, mass of the person is m = 55 kg . Hence force applied to the person is 2 5 F = ma = (55 kg)(4858 m/s ) = 2.67 × 10 N (c) If the duration of acceleration is given by t, then we have a = vu vu 1.3 m/s14 m/s 3 t = a = 4858 m/s = 2.6 × 10 s = 2.6 ms So the force last for 2.6 ms.

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Chapter 5, Problem 62P is Solved
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Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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