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Concept explainers
A particle moves in the xy plane (Fig. P9.30) from the origin to a point having coordinates x = 7.00 m and y = 4.00 m under the influence of a force given by →F=3y2î+x�.
- a. What is the work done on the particle by the force F if it moves along path 1 (shown in red)?
- b. What is the work done on the particle by the force F if it moves along path 2 (shown in blue)?
- c. What is the work done on the particle by the force F if it moves along path 3 (shown in green)?
- d. Is the force F conservative or nonconservative? Explain.
FIGURE P9.30
In each case, the work is found using the integral of →F·d→r along the path (Equation 9.21).
W=∫rfrt→F·d→r=∫rfrt(Fxdx+Fydy+Fzdz)
(a) The work done along path 1, we first need to integrate along d→r=dxˆi from (0,0) to (7,0) and then along d→r=dyˆj from (7,0) to (7,4):
W1=x=7;y=0∫x=0;y=0(3y2ˆi+xˆj)·(dxˆi)+x=7;y=4∫x=7;y=0(3y2ˆi+xˆj)·(dyˆj)
Performing the dot products, we get
W1=x=7;y=0∫x=0;y=03y2dx+x=7;y=4∫x=7;y=0xdy
Along the first part of this path, y = 0 therefore the first integral equals zero. For the second integral, x is constant and can be pulled out of the integral, and we can evaluate dy.
W1=0+x=7;y=4∫x=7;y=0xdy=xy|x=7;y=4x=7;y=0=28 J
(b) The work done along path 2 is along d→r=dyˆj from (0,0) to (0,4) and then along d→r=dxˆi from (0,4) to (7,4):
W2=x=0;y=4∫x=0;y=0(3y2ˆi+xˆj)·(dyˆj)+x=7;y=4∫x=0;y=4(3y2ˆi+xˆj)·(dyˆi)
Performing the dot product, we get:
W2=x=0;y=4∫x=0;y=0xdy+x=7;y=4∫x=0;y=43y2dx
Along the first part of this path, x = 0. Therefore, the first integral equals zero. For the second integral, y is constant and can be pulled out of the integral, and we can evaluate dx.
W2=0+3y2x|x=7;y=4x=0;y=4=336 J
(c) To find the work along the third path, we first write the expression for the work integral.
W=∫rfrt→F·d→r=∫rfrt(Fxdx+Fydy+Fzdz)W=∫rfrt(3y2dx+x dy) (1)
At first glance, this appears quite simple, but we can’t integrate ∫xdy=xy like we might have above because the value of x changes as we vary y (i.e., x is a function of y.) [In parts (a) and (b), on a straight horizontal or vertical line, only x or y changes]. One approach is to parameterize both x and y as a function of another variable, say t, and write each integral in terms of only x or y. Constraining dr to be along the desired line, we can relate dx and dy:
tanθ=dydxdy=tanθdx and dx=dytanθ (2)
Now, use equation (2) in (1) to express each integral in terms of only one variable.
W=x=7;y=4∫x=0;y=03y2dx+x=7;y=4∫x=0;y=0x dyW=y=4∫y=03y2dytanθ+x=7∫x=0x tanθ dx
We can determine the tangent of the angle, which is constant (the angle is the angle of the line with respect to the horizontal).
tanθ=4.007.00=0.570
Insert the value of the tangent and solve the integrals.
W=30.570y33|y=4y=0+0.570x22|x=7x=0W=112+14=126 J
(d) Since the work done is not “path-independent”, this is non-conservative force.
Figure P9.30ANS
(a)
![Check Mark](/static/check-mark.png)
The work done on the particle by the force F if it moves along the path 1.
Answer to Problem 30PQ
The work done on the particle by the force F if it moves along the path 1 is 28 J_.
Explanation of Solution
The path 1 followed by the particle is given in Figure P9.30. Given that the coordinate of the final position of the particle is (7.00, 4.00) and the force is →F=3y2ˆi+xˆj.
Write the expression for the work done by a force.
W=rf∫ri→F⋅d→r (I)
Here, W is the work done, rf is the final position, ri is the initial position, and d→r is the elemental path.
The path 1 of the particle consist of two parts. Motion from (0, 0) to (7, 0) and (7, 0) to (7, 4). Thus, the integration has to be performed along d→r=dxˆi from (0, 0) to (7, 0) and then along d→r=dyˆj from (7, 0) to (7, 4).
Use the force vector along with the limits of integration and perform the integration (represent the work done along path 1 as W1).
W1=x=7;y=0∫x=0;y=0(3y2ˆi+xˆj)·(dxˆi)+x=7;y=4∫x=7;y=0(3y2ˆi+xˆj)·(dyˆj) (II)
Perform the dot product to reduce the integral (II).
W1=x=7;y=0∫x=0;y=03y2dx+x=7;y=4∫x=7;y=0xdy (III)
Along the first part of the path 1, y=0, and hence the first integral equals zero. For the second integral, x is constant and can be pulled out of the integral, and dy can be evaluated.
W1=0+x=7;y=4∫x=7;y=0xdy=xy|x=7;y=4x=7;y=0=28−0=28 J
Conclusion:
Therefore, the work done on the particle by the force F if it moves along the path 1 is 28 J_.
(b)
![Check Mark](/static/check-mark.png)
The work done on the particle by the force F if it moves along the path 2.
Answer to Problem 30PQ
The work done on the particle by the force F if it moves along the path 2 is 336 J_.
Explanation of Solution
The path 2 followed by the particle is given in Figure P9.30. Given that the coordinate of the final position of the particle is (7.00, 4.00) and the force is →F=3y2ˆi+xˆj.
Equation (I) gives the expression for the work done by a force.
W=rf∫ri→F⋅d→r
The path 2 of the particle consist of two parts. Motion from (0, 0) to (0, 4) and (0, 4) to (7, 4). Thus, the integration has to be performed along d→r=dyˆj from (0, 0) to (0, 4) and then along d→r=dxˆi from (0, 4) to (7, 4).
Use the force vector along with the limits of integration and perform the integration (represent the work done along path 2 as W2).
W2=x=0;y=4∫x=0;y=0(3y2ˆi+xˆj)·(dyˆj)+x=7;y=4∫x=0;y=4(3y2ˆi+xˆj)·(dxˆi) (IV)
Perform the dot product to reduce the integral (IV).
W2=x=0;y=4∫x=0;y=0xdy+x=7;y=4∫x=0;y=43y2dx (V)
Along the first part of the path 2, x=0, and hence the first integral equals zero. For the second integral, y is constant and can be pulled out of the integral, and dx can be evaluated.
W2=0+x=7;y=4∫x=0;y=43y2dx=3y2x|x=7;y=4x=0;y=4=336−0=336 J
Conclusion:
Therefore, the work done on the particle by the force F if it moves along the path 2 is 336 J_.
(c)
![Check Mark](/static/check-mark.png)
The work done on the particle by the force F if it moves along the path 3.
Answer to Problem 30PQ
The work done on the particle by the force F if it moves along the path 3 is 126 J_.
Explanation of Solution
The path 3 followed by the particle is given in Figure P9.30. Given that the coordinate of the final position of the particle is (7.00, 4.00) and the force is →F=3y2ˆi+xˆj.
Equation (I) gives the expression for the work done by a force.
W=rf∫ri→F⋅d→r
Write equation (I) in terms of x, y, and z components.
W=rf∫ri(Fxdx+Fydy+Fzdz) (VI)
Use the x and y components of the given force in the integral (VI). Since the motion is confined in xy plane, there is no z component in the integral.
W=rf∫ri(3y2dx+xdy) (VII)
The path 3 of the particle starts from (0, 0) and ends at (7, 4). Thus, the integral (VII) can be written as (represent the work done along path 3 as W3),
W3=x=7;y=4∫x=0;y=03y2dx+x=7;y=4∫x=0;y=0xdy (VIII)
Here, both x and y coordinates change and hence direct integration cannot be performed. Let us consider the angle θ that the vector of path 3 make with the x axis as shown in Figure 1.
Write the expression relating dx, dy, and θ.
tanθ=dydx (IX)
Solve equation (IX) for dy.
dy=tanθdx (X)
Solve equation (IX) for dx.
dx=dytanθ (XI)
Use equation (X) and (XI) in (VIII).
W3=x=7;y=4∫x=0;y=03y2dytanθ+x=7;y=4∫x=0;y=0xtanθdx (XII)
Compute tanθ from the paths shown in Figure P9.30.
tanθ=4.007.00=0.570 (XIII)
Use equation (XIII) in (XII) and perform the integral.
W3=y=4∫y=030.570y2dy+x=7∫x=00.570xdx=30.570y33|y=4y=0+0.570x22|x=7x=0=112+14=126 J
Conclusion:
Therefore, the work done on the particle by the force F if it moves along the path 3 is 126 J_.
(d)
![Check Mark](/static/check-mark.png)
Whether the force →F is conservative or non-conservative.
Answer to Problem 30PQ
The force →F is non-conservative in nature.
Explanation of Solution
From part (a), (b) and (c) it is found that the work done by the force →F on moving the particle from (0,0) to (7, 4) along different path is different. By the definition of conservative forces, the work done by a conservative force is path independent. However, here the work done by the force →F is path dependent and hence it is a non-conservative force.
Conclusion:
Therefore, the force →F is non-conservative in nature.
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Chapter 9 Solutions
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