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If you tried to smuggle gold bricks by filling your

Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli ISBN: 9780130606204 3

Solution for problem 3P Chapter 10

Physics: Principles with Applications | 6th Edition

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Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli

Physics: Principles with Applications | 6th Edition

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Problem 3P

Problem 3P

If you tried to smuggle gold bricks by filling your backpack, whose dimensions are 60 cm × 28 cm × 18 cm, what would its mass be?

Step-by-Step Solution:
Step 1 of 3

ANSWER: STEP 1:- For the above problem, the acceleration due to gravity value has taken as g = 10 m/s .2 The downward force which is 2500 N. The displacement is 5 meters. Work done by the force is, W = F. ds = F. S = |F||S|cos As the force is independent of the displacement here or constant, we took it out of the integral sign and the integral over the infinitesimal displacement vector will become the total displacement. 0 The angle between force and displacement here is 0 , as both are in the same direction. Then work done will be, W = 2500 N × 5 m × cos 0 = 2500 N × 5 m × 1 = 12500 jules. STEP 2:- 0 The tension T m1kes an angle 60 the horizontal, so the angle with the displacement which is vertically downwards will be, 0 0 0 60 + 90 = 150 . The magnitude of force T is11830 N. So, the work done by the force T wi1l be, W = F. ds = F. S = |F||S|cos As the force is independent of the displacement here or constant, we took it out of the integral sign and the integral over the infinitesimal displacement vector will become the total displacement. The angle between force and displacement here is 150 . 0 Then work done will be, W = 1830 N × 5 m × cos 150 = 1830 N × 5 m × ( 0.866). = 7924.13 jules The minus sign is trivial, as the cosine of 150 is negative. So the work is negative. STEP 3:- 0 The tension T m2kes an angle 45 the horizontal, so the angle with the displacement which is vertically downwards will be, 0 0 0 45 + 90 = 135 . The magnitude of force T is 1295 N. So, the work done by the force T wi2l be, W = F. ds = F. S = |F||S|cos As the force is independent of the displacement here or constant, we took it out of the integral sign and the integral over the infinitesimal displacement vector will become the total displacement. 0 The angle between force and displacement here is 150 . Then work done will be, 0 W = 1295 N × 5 m × cos 135 = 1295 N × 5 m × ( 0.707). = 4578.51 jules The minus sign is trivial, as the cosine of 135 is negative. So the work is negative. CONCLUSION: The work done by the 3 forces we got as, 12500 jules, 7924.13 jules and 4578.51 jules.

Step 2 of 3

Chapter 10, Problem 3P is Solved
Step 3 of 3

Textbook: Physics: Principles with Applications
Edition: 6
Author: Douglas C. Giancoli
ISBN: 9780130606204

The full step-by-step solution to problem: 3P from chapter: 10 was answered by , our top Physics solution expert on 03/03/17, 03:53PM. This full solution covers the following key subjects: done, exactly, figure, forces, Ground. This expansive textbook survival guide covers 35 chapters, and 3914 solutions. Since the solution to 3P from 10 chapter was answered, more than 551 students have viewed the full step-by-step answer. The answer to “If you tried to smuggle gold bricks by filling your backpack, whose dimensions are 60 cm × 28 cm × 18 cm, what would its mass be?” is broken down into a number of easy to follow steps, and 27 words. Physics: Principles with Applications was written by and is associated to the ISBN: 9780130606204. This textbook survival guide was created for the textbook: Physics: Principles with Applications, edition: 6.

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