Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

A 20 kg 0object is acted on by a conservative force given by $F = - 3.0 x - 5.0 x^{2}$ , with F in newton and x in meters. Take the potential energy associated with the force to be zero when the object is at x=0. (a) What is the potential energy of the system associated with the force when the object is at x=2.0 m? (b) If the object has a velocity of 4.0 m/s in the negative direction of the x-axis when it is at x=5.00 m, what is its speed when it passes through the origin? (c) What are the answers to (a) and (b) if the potential energy of the system is taken to be -8.0 J when the object is at x=0?

a) The P.E. of the system associated with the force when the object is at x=2.0 m is 19.4 J

b) The speed of the object when it passes through the origin is 6.37 m/s

c) Answer for part a) will be 11.4 J and that for part b) will be 6.37 m/s when P.E. of the system when the object is at x=0 is-8.0 J.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given data

The mass of an object is m=20 kg

The force applied to an object is $F = - 3.0 x - 5.0 x^{2}$

The potential energy of the object at x=0 is, $U_{0} = 0 J$

The velocity of an object is v=4.0 m/s in the negative direction of the x-axis when it is at x=0

Step 2: Understanding the concept of work done

Using the formula for work done on a system by an external force, we can find the stopping distance of an automobile.

Formulae:

The potential energy or work done by the system, $U (x) = \int F (x) d x$ (1)

The total energy of the system, $E = \frac{1}{2} m v^{2} + U (x)$ (2)

From the law of conservation of energy,

$E_{i} = E_{f}$ (3)

Step 3: a) Calculation of the potential energy of the system

We are given that the force applied to an object is $F = - 3.0 x - 5.0 x^{2}$

Then P.E associated with it is using equation (1) is given as:

$U (x) = \int_{x i}^{x f} - 3.0 x - 5.0 x^{2} d x = (- \frac{3.0 x^{2}}{2} - \frac{5.0 x^{3}}{3}) - U_{0} = (\frac{3.0 x^{2}}{2} + \frac{5.0 x^{3}}{3}) + U_{0}$

$|U (x = 2.0 m)| = (\frac{3.0 {(2.0 m)}^{2}}{2} + \frac{5.0 {(2.0 m)}^{3}}{3}) + 0 = 19.4 J$

Therefore, the P.E. of the system associated with the force when the object is at x=2.0 m is equal to 19.4J

Step 4: b) Calculation of the speed of the object

But, it has the speed of $v = 4.0 \frac{m}{s} at x = 5.0 m$

From equation (3), we can get the initial speed of the object as follows:

$E_{i} (x = 0 m) = E_{f} (x = 5 m) \frac{1}{2} m v_{0}^{2} + U_{0} = \frac{1}{2} m v^{2} + U (x = 5 m) \frac{1}{2} (20 k g) v_{0}^{2} + 0 \frac{1}{2} (20 k g) {(4.0 \frac{m}{s})}^{2} + (\frac{3.0 {(5 m)}^{2}}{2} + \frac{5.0 {(5 m)}^{3}}{3} + 0) v_{0}^{2} = 40.58 \frac{m^{2^{2}}}{s} v_{0} = 6.37 \frac{m}{s}$

Therefore, the speed of an object when it passes through the origin is 6.37 m/s

Step 5: c) Calculation of the parts (a) and (b) for potential energy -8.0J

From part a), we can write that,

$|U (x = 2 m)| = (\frac{3.0 {(2 m)}^{2}}{2} + \frac{5.0 {(2 m)}^{3}}{3}) + U_{0}$

If $U_{0} = - 8 J$ then, the new potential energy is given as:

$|U (x = 2 m)| = (\frac{3.0 {(2 m)}^{2}}{2} + \frac{5.0 {(2 m)}^{3}}{3}) - 8 J = 11.4 J$

$v_{0}$ will be the same as before because $U_{0}$ on both sides of the equation part (b) cancels out.

So it is independent of $U_{0}$ .

Therefore, the answer for part a) will be 11.4 J and that for part b) will be 6.37 m/s.

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of work done

Step 3: a) Calculation of the potential energy of the system

Step 4: b) Calculation of the speed of the object

Step 5: c) Calculation of the parts (a) and (b) for potential energy -8.0J

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Fundamentals Of Physics

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Step 1: The given data

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Step 4: b) Calculation of the speed of the object

Step 5: c) Calculation of the parts (a) and (b) for potential energy -8.0J

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