Home»Math Guides»Work done by vector field onto particle
Work Done By Force Field on Particle (Vector Fields) Example 1
Find the work done by the force field \(F(x,y) = {x^2}\mathop i\limits^ \to + xy\mathop j\limits^ \to \) on a particle that moves once counterclockwise around the circle governed by \({x^2} + {y^2} = 4\)
We need to make a vector equation for “r”, which will tell us the path this goes in, and we should use polar coordinates since we see we’re given a circle.
We’re told the path is a circle of radius 2 (square root the 4).
In other words, \(r(t) = \left\langle {2\cos (t),2\sin (t)} \right\rangle \) for \(0 \le t \le 2\pi \)
Why? Treat the components in r(t) as x and y, plug the x component into x squared, and plug the y component into y squared, and you get 4 = 4 as a check.
We’re also told “once counterclockwise”, so it starts from 0 to \(2\pi \) since each quadrant is \(\frac{\pi }{2}\).
We need to find the derivative of r(t).
\[r'(t) = \left\langle { – 2\sin (t),2\cos (t)} \right\rangle \]
Then we find F(r(t)), which is just substituting our r(t) into F(x,y).
\[F\left( {r(t)} \right) = \left\langle {4{{\cos }^2}(t),4\sin (t)\cos (t)} \right\rangle \]
Then all we need to find is \(\int_0^{2\pi } {\left( {F \cdot dr} \right)} dt\).
This is just the dot product between F(r(t)) and r'(t) and we use our domain (once counterclockwise means 0 to 2pi) as the integral boundaries.
From there it’s algebra and trigonometry in the integral.
\[\int_0^{2\pi } {\left( {F \cdot dr} \right)} dt\]
\[\int_0^{2\pi } {\left( {4{{\cos }^2}(t)\left( { – 2\sin (t)} \right) + 4\sin (t)\cos (t) \cdot 2\cos (t)} \right)} dt\]
\[\int_0^{2\pi } {\left( { – 8{{\cos }^2}(t)\sin (t) + 8{{\cos }^2}(t)\sin (t)} \right)} dt\]
\[ = 0\]
So the work done by the force field is 0.
Why would this happen?
The vectors in the field are simply perpendicular to the path r(t), it’s never pushing/pulling the particular in the direction it’s going in.
Try another example where you find work done by vector field onto particle