Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

A 12.0 N force with a fixed orientation does work on a particle as the particle moves through the three-dimensional displacement $\vec{d} = (2.00 \hat{i} - 4.00 \hat{j} + 30.0 \hat{k})$ . What is the angle between the force and the displacement if the change in the particle’s kinetic energy is (a) +30.0 J and (b) -30.0 J ?

The angle between the force and the displacement is $ϕ = 62 . 3^{°}$ .
The angle between the force and the displacement is $ϕ = 118^{°}$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data:

The force, $F = 12 N$

The displacement, $\vec{d} = (2.00 \hat{i} - 4.00 \hat{j} + 30.0 \hat{k})$

Step 2: Understanding the concept:

The work-energy theorem states that the net work done by forces on an object is equal to the change in its kinetic energy.

Using the work-kinetic energy theorem, you have

$\begin{array}{rcl} ∆ K & = & W \\ = & \vec{F} . \vec{d} \\ = & Fdcosϕ \end{array}$

Step 3: The magnitude of the displacement:

Calculate the magnitude of the displacement as below.

$\begin{array}{rcl} d & = & \sqrt{{(2.00 m)}^{2} + {(- 4.00 m)}^{2} + {(3.00 m)}^{2}} \\ = & \sqrt{(4 + 16 + 9)} m \\ = & \sqrt{29} m \\ = & 5.39 m \end{array}$

Step 4: (a) The angle between the force and the displacement if the kinetic energy is +30.0 J :

The angle between the force and the displacement if the change in the particle’s kinetic energy is,

$∆ K = + 30.0 J$

Therefore, the angle is,

$\begin{array}{rcl} ϕ & = & \cos^{- 1} (\frac{∆ K}{Fd}) \\ = & \cos^{- 1} (\frac{30.0 J}{(12.0 N) (5.39 m)}) \\ = & \cos^{- 1} (0.464) \\ = & 62 . 3^{°} \end{array}$

Hence, the angle between the force and the displacement is $62 . 3^{°}$ .

Step 5: (b) The angle between the force and the displacement if the kinetic energy is -30.0 J :

The angle between the force and the displacement if the change in the particle’s kinetic energy is,

$∆ K = - 30.0 J$

Therefore, the angle is,

$\begin{array}{rcl} ϕ & = & \cos^{- 1} (\frac{∆ K}{Fd}) \\ = & \cos^{- 1} (\frac{- 30.0 J}{(12.0 N) (5.39 m)}) \\ = & \cos^{- 1} (- 0.464) \\ = & 117 . 64^{°} \\ ϕ & \approx & 118^{°} \end{array}$

Hence, the angle between the force and the displacement is $118^{°}$ .

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Given data:

Step 2: Understanding the concept:

Step 3: The magnitude of the displacement:

Step 4: (a) The angle between the force and the displacement if the kinetic energy is +30.0 J :

Step 5: (b) The angle between the force and the displacement if the kinetic energy is -30.0 J :

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Fundamentals Of Physics

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Short Answer

A 12.0 N force with a fixed orientation does work on a particle as the particle moves through the three-dimensional displacement d→=(2.00i^-4.00j^+30.0k^). What is the angle between the force and the displacement if the change in the particle’s kinetic energy is (a) +30.0 J and (b) -30.0 J ?

Step by Step Solution

Step 1: Given data:

Step 2: Understanding the concept:

Step 3: The magnitude of the displacement:

Step 4: (a) The angle between the force and the displacement if the kinetic energy is +30.0 J :

Step 5: (b) The angle between the force and the displacement if the kinetic energy is -30.0 J :

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