Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

A proton $(mass m = 1.67 \times 10^{- 27} kg)$ is being accelerated along a straight line at $3.6 \times 10^{15} m / s^{2}$ in a machine. If the proton has an initial speed of $2.4 \times 10^{7} m / s$ and travels 3.5 cm, what then is (a) its speed and (b) the increase in its kinetic energy?

The final speed of the proton is $2.9 \times 10^{7} m / s$ .
The increase in the kinetic energy of the proton is $2.1 \times 10^{- 13} J$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

Mass of the proton is, $m = 1.67 \times 10^{- 27} kg$ .
Acceleration is, $a = 3.6 \times 10^{15} m / s^{2}$ .
The initial speed is, $v_{i} = 2.4 \times 10^{7} m / s$ .
Distance traveled is, $d = 3.5 cm = 0.035 m$ .

Step 2: Understanding the concept

Using the third kinematic equation, we can find the final speed of the proton from a given mass, acceleration, and initial speed of the proton. Also, we can use this final velocity and given initial velocity to calculate the increase in its kinetic energy.

Step 3: a) Calculate the speed of the proton

From newtons law of motion equations, the final velocity can be calculated as,

$v_{f}^{2} = v_{i}^{2} + 2 ad$

Substitute the given values in the above equation, and we get,

$v_{f}^{2} = {(2.4 \times 10^{7} \frac{m}{s})}^{2} + 2 (3.6 \times 10^{15} \frac{m}{s^{2}}) (0.035 m) v_{f}^{2} = 5.76 \times 10^{14} m^{2} / s^{2} + 2.52 \times 10^{14} m^{2} / s^{2} v_{f}^{2} = 8.41 \times 10^{14} m^{2} / s^{2} v_{f} = 2.9 \times 10^{7} m / s$

Therefore, the final speed of the proton is $2.9 \times 10^{7} m / s$ .

Step 4: (b) Calculate the increase in the proton’s kinetic energy

The expression for kinetic energy is,

$K = \frac{1}{2} {mv}^{2}$

Initial kinetic energy can be calculated as,

$K_{i} = \frac{1}{2} {mv}_{i}^{2}$

Substitute the given values in the above equation, and we get,

$\begin{array}{rcl} K_{i} & = & \frac{1}{2} (1.67 \times 10^{- 27} kg) {(2.4 \times 10^{7} \frac{m}{s})}^{2} \\ = & 4.8 \times 10^{- 13} . (1 kg \times 1 \frac{m^{2}}{s^{2}} \times \frac{1 J}{1 N . m} \times \frac{1 N}{1 kg . \frac{m}{s^{2}}}) \\ = & 4.8 \times 10^{- 13} J \end{array}$

Final kinetic energy can be calculated as,

$K_{f} = \frac{1}{2} {mv}_{f}^{2}$

Substitute the given values in the above equation, and we get,

$\begin{array}{rcl} K_{f} & = & \frac{1}{2} (1.6 \times 10^{- 27} kg) {(2.9 \times 10^{7} \frac{m}{s})}^{2} \\ = & 6.9 \times 10^{- 13} . (1 kg \times 1 \frac{m^{2}}{s^{2}} \times \frac{1 J}{1 N . m} \times \frac{1 N}{1 kg . \frac{m}{s^{2}}}) \\ = & 6.9 \times 10^{- 13} J \end{array}$

Therefore, an increase in the kinetic energy of the proton can be calculated as,

$∆ K = K_{f} - K_{i}$

Substitute the given values in the above equation, and we get,

$∆ K = 6.9 \times 10^{- 13} J - 4.8 \times 10^{- 13} J ∆ K = 2.1 \times 10^{- 13} J$

Therefore, an increase in the kinetic energy of the proton is $2.1 \times 10^{- 13} J$ .

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Fundamentals Of Physics

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Short Answer

A proton $(mass m = 1.67 \times 10^{- 27} kg)$ is being accelerated along a straight line at $3.6 \times 10^{15} m / s^{2}$ in a machine. If the proton has an initial speed of $2.4 \times 10^{7} m / s$ and travels 3.5 cm, what then is (a) its speed and (b) the increase in its kinetic energy?

Step by Step Solution

Step 1: Given data

Step 2: Understanding the concept

Step 3: a) Calculate the speed of the proton

Step 4: (b) Calculate the increase in the proton’s kinetic energy

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Fundamentals Of Physics

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Short Answer

A proton (mass m=1.67×10-27 kg) is being accelerated along a straight line at 3.6×1015 m/s2 in a machine. If the proton has an initial speed of 2.4×107 m/s and travels 3.5 cm, what then is (a) its speed and (b) the increase in its kinetic energy?

Step by Step Solution

Step 1: Given data

Step 2: Understanding the concept

Step 3: a) Calculate the speed of the proton

Step 4: (b) Calculate the increase in the proton’s kinetic energy

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A proton $(mass m = 1.67 \times 10^{- 27} kg)$ is being accelerated along a straight line at $3.6 \times 10^{15} m / s^{2}$ in a machine. If the proton has an initial speed of $2.4 \times 10^{7} m / s$ and travels 3.5 cm, what then is (a) its speed and (b) the increase in its kinetic energy?