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CP In a shipping company distribution center, an open cart

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

Solution for problem 95P Chapter 8

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 95P

CP? In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s (?Fig. P8.87?). Ignore friction between the cart and the floor. A 15.0-kg package slides down a chute that is inclined at 37o from the horizontal and leaves the end of the chute with a speed of 3.00 m/s. The package lands in the cart and they roll together. If the lower end of the chute is a vertical distance of 4.00 m above the bottom of the cart, what are (a) the speed of the package just before it lands in the cart and (b) the final speed of the cart?

Step-by-Step Solution:
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Solution 95P This question can be solved considering the equations of conservation of energy and momentum. Given, mass of the package m = 15p0 kg Speed of the package v = 3p00 m/s Height of the chute above the bottom of the cart h = 4.00 m Let v b1 the speed of the package when it lands in the cart. Therefore, using the principle of conservation of energy, (a) Total initial energy = Total final energy 1 2 1 2 2m vp p + m gp = m v2 p 1 2 2 v1 = v p + 2gh v1 2 = (3.00) + 2 × 9.8 × 4 m /s2 2 2 2 2 v1 = 9.00 + 78.4 m /s v1= 9.35 m/s The speed of the package when it lands in the cart is 9.35 m/s. (b) The cart is moving to the left with speed of 5.00 m/s. So, v =c 5.00 m/s Mass of the cart m = c0.0 kg 0 The package slides down the inclined plane. The angle of inclination is 37 C . 0 Therefore, its x-component of speed is v px= 3.00 m/s × cos 37 = 2.39 m/s Initial momentum of the system, P = 15.0 kg × 2.39 m/s + 50.0 kg × ( 5.00 m/s) = 214.15 kg.m/s i If the final speed of the cart is v2. Final momentum of the system P = (15 + 50)v = 65v f 2 2 Therefore, P = Pi f 214.15 kg.m/s = 65v 2 v =2 3.29 m/s The speed of the cart would be 3.29 m/s. The negative sign indicates that the cart would move to the left.

Step 2 of 3

Chapter 8, Problem 95P is Solved
Step 3 of 3

Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

The full step-by-step solution to problem: 95P from chapter: 8 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. University Physics was written by and is associated to the ISBN: 9780321675460. This full solution covers the following key subjects: cart, speed, package, chute, lands. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. Since the solution to 95P from 8 chapter was answered, more than 543 students have viewed the full step-by-step answer. The answer to “CP? In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s (?Fig. P8.87?). Ignore friction between the cart and the floor. A 15.0-kg package slides down a chute that is inclined at 37o from the horizontal and leaves the end of the chute with a speed of 3.00 m/s. The package lands in the cart and they roll together. If the lower end of the chute is a vertical distance of 4.00 m above the bottom of the cart, what are (a) the speed of the package just before it lands in the cart and (b) the final speed of the cart?” is broken down into a number of easy to follow steps, and 116 words. This textbook survival guide was created for the textbook: University Physics, edition: 13.

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CP In a shipping company distribution center, an open cart