Fundamentals of Physics Extended
Fundamentals of Physics Extended
10th Edition
ISBN: 9781118230725
Author: David Halliday, Robert Resnick, Jearl Walker
Publisher: Wiley, John & Sons, Incorporated
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Chapter 7, Problem 1Q

Rank the following velocities according to the kinetic energy a particle will have with each velocity, greatest first: (a) v = 4 i ^ + 3 j ^ , (b) v = −4 i ^ + 3 j ^ , (c) v = −3 i ^ + 4 j ^ , (d) v = 3 i ^ − 4 j ^ , (e) v = 5 i ^ , and (f) v = 5 m/s at 30° to the horizontal.

Expert Solution & Answer
Check Mark
To determine

To rank:

The velocity according to the greatest kinetic energy at first.

Answer to Problem 1Q

Solution:

All velocities have the same kinetic energy.

Explanation of Solution

1) Concept:

To rank kinetic energy, we have to find first find the resultant velocity, and then use the formula for kinetic energy in which the mass is the same for all velocities. Therefore, kinetic energy is ranked only on the magnitude of velocity.

2) Formula:

i) v=vx2+vy2

ii) KE=0.5×m×v2

3) Given:

i) v=4i^+3j^

ii) v=-4i^+3j^

iii) v=-3i^+4j^

iv) v=-3i^-4j^

v) v=5i^

vi) v=5   at  30° to horizontal v=5cos30i^+5sin30j^

4) Calculation:

First find the resultant velocity as follows:

v=vx2+vy2

For v=4i^+3j^

v=42+32

v=5 m/s

For v=-4i^+3j^

v=(-4)2+32

v=5 m/s

For v=-3i^+4j^

v=-32+42

v=5 m/s

For v=-3i^-4j^

v=-32+(-42)

v=5 m/s

For v=5i^

v=52

v=5 m/s

For v=5cos30i^+5sin30j^

v=52(cos230+sin230)

v=5 m/s

Now kinetic energy can be found by using the following formula:

KE=0.5×m×v2

Assume m=1 kg

So

KE=0.5×v2

As the resultant velocity  v=5m/s is the same for all cases, kinetic energy is also the same for all.

Conclusion:

Resultant velocity is found by using Pythagoras theorem; from where the kinetic energy could be calculated.

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Chapter 7 Solutions

Fundamentals of Physics Extended

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  • A block of mass 0.500 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x (Fig. P7.79). The force constant of the spring is 450 N/m. When it is released, the block travels along a frictionless, horizontal surface to point , the bottom of a vertical circular track of radius R = 1.00 m, and continues to move up the track. The blocks speed at the bottom of the track is = 12.0 m/s, and the block experiences an average friction force of 7.00 N while sliding up the track. (a) What is x? (b) If the block were to reach the top of the track, what would be its speed at that point? (c) Does the block actually reach the top of the track, or does it fall off before reaching the top?
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