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A variable force $\vec{F} = 3x^2y \vec{i}$ (unit vector) is acting on an object that travels on a square loop in a clockwise direction starting from the origin to (0,5) then (5,5) then (5,0) then back to (0,0). How much work is done on the particle by the force during one complete trip around the square?

It should be 625 Joules because when the particle is on the $x$-axis, $x=0$ so there's no force, and when the particle is moving up or down the force is not in its direction?

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  • $\begingroup$ Are you sure that your units are right? is the force in newtons and the coordinates in meters (for example)? if it is so then the answer should be 625 J. $\endgroup$ – 4nt Oct 19 '15 at 4:26
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This is like calculating a line integral only.
Work done, $W= \oint_C \vec F\cdot d\vec r = \oint_C 3x^2ydx=\int_I 3x^2ydx +\int_{II} 3x^2ydx +\int_{III} 3x^2ydx +\int_{IV} 3x^2ydx$
Now for path I, (0,0) to (0,5): $dx = 0$ $$\int_I 3x^2ydx = 0$$ Now for path II, (0,5) to (5,5): $y=5$ $$\int_{II} 3x^2ydx = 15\int^5_0 x^2 dx=5(5^3-0) = 625$$ Now for path III, (5,5) to (5,0): $dx = 0$ $$\int_{III} 3x^2ydx = 0$$ Now for path IV, (5,0) to (0,0): $y=0$ $$\int_{IV} 3x^2ydx = 0$$

Hence we can say that net work done = $W= \oint_C \vec F\cdot d\vec r = \oint_C 3x^2ydx = 625 J$

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  • $\begingroup$ But 625 Joules is not the answer according to the physics solution? $\endgroup$ – Goldname Oct 20 '15 at 0:21
  • $\begingroup$ What is the answer according to the solution? $\endgroup$ – SchrodingersCat Oct 20 '15 at 14:01
  • $\begingroup$ It is -188 Joules $\endgroup$ – Goldname Oct 20 '15 at 20:10
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Your force $F=3x^2y\hat{i}$ has only x component suggest that it is acting only in x direction. So, motion along y axis does not cause any work done by the force, as it is perpendicular to the direction of force. Work is said to be done only along x axis . So,
$$W= F.S\\ = \int_{_(0,5)}^{(5,5)}{f.dx} + \int_{_(5,0)}^{(0,0)}{f.dx} = {5^3}\into{5}-0 = 625 unit$$

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