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A 350-kg roller coaster car starts from rest at point A

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

Solution for problem 45P 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 45P

A 350-kg roller coaster car starts from rest at point A and slides down a frictionless loop-the-loop (Fig. P7.41). (a) How fast is this roller coaster car moving at point B? (b) How hard does it press against the track at point B?

Step-by-Step Solution:

Solution 45P Step 1: Provided that, the loop is frictionless. So, there will not be any loss is energy due to frictional force. Initially, the roller coaster car will possess only kinetic energy, since it is at rest. We can represent it as U A Therefore, U =Agh Where, m - mass of the roller coaster car g - Acceleration due to gravity h - initial height of the roller coaster car from the ground 2 Provided, m = 350 kg, g = 9.8 m/s and h = 25.0 m 2 Therefore, U = A0 kg × 9.8 m/s × 25 m = 85750 J Similarly, the potential energy at point B is, U = mgh’ B Where, h’ - Height of point B from the ground 2 U A350 kg × 9.8 m/s × 12 m = 41160 J Step 2: a) We can see a change in potential energy in between points A and B That is, U -A = B750 J - 41160 J = 44590 J This much energy is converted to the kinetic energy of the roller coaster car. So we can write, ½ mv = 44590 J Where, v - speed of the roller coaster car. That is, ½ × 350 kg × v = 44590 J v = 4 4590 J × 2 / 350 kg = 254.8 m /s 2 2 Taking square root on both sides we get, v = 15.96 m/s The speed of the roller coaster car at point B will be, v = 15.96 m/s

Step 3 of 3

Chapter 7, Problem 45P is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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