Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Textbook Question
Chapter 7, Problem 7.46P

An object moves in the xy plane 111 Figure P7.43 and experiences a friction force with constant magnitude 3.00 N, always acting in the direction opposite the object's velocity. Calculate the work that you must do to slide the object at constant speed against the friction force as the object moves along (a) the purple path O to Ⓐ followed by a return purple path to O, (b) the purple path O to © followed by a return blue path to O, and © the blue path O to © followed by a return blue path to O. (d) Each of your three answers should be nonzero. What is the significance of this observation?

Chapter 7, Problem 7.46P, An object moves in the xy plane 111 Figure P7.43 and experiences a friction force with constant

(a)

Expert Solution
Check Mark
To determine

The work done by the frictional force on the particle as the particle moves along purple path O to A followed by a return purple path to O .

Answer to Problem 7.46P

The work done by the frictional force on the particle is 30J .

Explanation of Solution

Given info: The friction force with constant magnitude is 3.00N and the position of particle is 5.00m along the path OA .

The particle moves along the x -axis and force is in the direction of the particle’s velocity so the angle between force and position of the particle is 0° .

The expression for the work done of the particle along the path OA is,

WOA=Fdx=Fxcosθ

Here,

F is the friction force acts in the opposite direction of the particle’s velocity.

x is the position of the particle along x -axis.

θ is the angle between force vector and position vector.

Substitute 3.00N for F , 5.00m for x and 0° for θ in the above equation.

WOA=(3.00N)(5.00m)cos0°=(3.00N)(5.00m)(1)=15J

The expression for the work done of the particle along the return path AO is,

WAO=Fdx=Fxcosθ

Substitute 3.00N for F , 5.00m for x and 0° for θ in the above equation.

WAO=(3.00N)(5.00m)cos0°=(3.00N)(5.00m)(1)=15J

Total work done by the frictional force along the path OA and return path AO is,

W1=WOA+WAO

Substitute 15J for WOA and 15J for WAO in the above equation.

W1=15J+15J=30J

Conclusion:

Therefore, the work done by the frictional force on the particle is 30J .

(b)

Expert Solution
Check Mark
To determine

The work done by the frictional force on the particle as the particle moves along purple path O to C followed by a return purple path to O .

Answer to Problem 7.46P

The work done by the frictional force on the particle is 51.21J .

Explanation of Solution

The position of particle is 5.00m along the path AC .

From part (a), the work done of the particle along the path OA is,

WOA=15J

The expression for the work done of the particle along the path AC is,

WAC=Fdy=Fycosθ

Here,

y is the position of particle along y -axis.

Substitute 3.00N for F , 5.00m for y and 0° for θ in the above equation.

WAC=(3.00N)(5.00m)cos0°=(3.00N)(5.00m)(1)=15J

The expression for the work done of the particle along the path CO is,

WCO=Fdr=Frcosθ (1)

Here,

r is the position of the particle along xy plane.

From the Pythagoras theorem, the value of r is,

r=(OA)2+(AC)2

Substitute 5.00m for OA and 5.00m for AC in the above equation.

r=(5.00m)2+(5.00m)2=52m

Substitute 3.00N for F , 52m for r and 0° for θ in the equation (1).

WCO=(3.00N)(52m)cos0°=(3.00N)(52m)(1)21.2132J

The total work done of the particle along the path OACO is,

W2=WOA+WAC+WCO

Substitute 15J for WOA , 15J for WAC and 21.2132J for WCO in the above equation.

W2=15J+(15J)+(21.2132J)=51.2132J51.21J

Conclusion:

Therefore, the work done by the frictional force on the particle is 51.21J .

(c)

Expert Solution
Check Mark
To determine

The work done by the frictional force on the particle as the particle moves along blue path O to C followed by a return blue path to O .

Answer to Problem 7.46P

The work done by the frictional force on the particle is 42.42J .

Explanation of Solution

From part (b), the work done of the particle along the return path CO is,

WCO=21.2132J

The work done of the particle along the path OC is,

WOC=21.2132J

Total work done by the frictional force along the path OCO is,

W3=WOC+WCO

Substitute 21.2132J for WOC and 21.2132J for WCO in the above equation.

W3=21.2132J+21.2132J=42.4264J42.42J

Conclusion:

Therefore, the work done by the frictional force on the particle is 42.42J .

(d)

Expert Solution
Check Mark
To determine

The significance of the observation when all three answers should be nonzero.

Answer to Problem 7.46P

The frictional force is a non-conservative force.

Explanation of Solution

The expression for the work done by the frictional force on the particle is,

WOA=Fdx=Fxcosθ

From part (a), the work done of the particle along the path OAO is,

W1=30J

From part (b), the work done of the particle along the path OACO is,

W2=51.21J

From part (b), the work done of the particle along the path OCO is,

W3=42.42J

The above equations shows, the value of work done by the frictional force on the particle in all closed paths and that are non-zero because the particle pushes with a force 3.00N in the direction of particle’s velocity that give the angle 0° between the force vector and position vector.

The value of cos0° is 1 that give the non-zero values of work done in all closed paths so the friction is a non-conservative force.

Conclusion:

Therefore, the frictional force is a non-conservative force.

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Chapter 7 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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