Find the total work done in moving a particle in a force field given by:

$F=(y^2-x^2)i+(2-x+y)j$ along the curve $y=x^3$, from $(-1,-1)$ to $(1,1)$

Help is appreciated!

  • $\begingroup$ What is your definition of "work"? $\endgroup$ – Joe Johnson 126 Oct 15 '13 at 15:39
  • $\begingroup$ Thanks for editing my question! I rewrote the question, is it any clearer now what I am looking for? $\endgroup$ – Soakr Oct 15 '13 at 15:46
  • $\begingroup$ I know what you are looking for. I am just wondering which definition of "work" to use in my answer. What is the definition of "work" that you use? Include it in the statement of your problem. $\endgroup$ – Joe Johnson 126 Oct 15 '13 at 15:47
  • $\begingroup$ I believe 'my' definition of the work of F would be the integral of work elements, dW. Is that it? I must say that I am probably a bit unclear on all the definitions :). $\endgroup$ – Soakr Oct 15 '13 at 16:17
  • $\begingroup$ More elementary I would say that I am looking for the energy that the field adds to the particle ;). $\endgroup$ – Soakr Oct 15 '13 at 16:22

In the case of a constant force and a linear motion, work is given by finding the component of the force acting in the direction of motion and then multiplying that by the distance over which you travel. Things are more difficult when you're moving in the plane.

Let ${\bf x}(t)$ give the path along which you move, then $\dot{\bf x}(t) = {\bf v}(t)$ gives the direction of (instantaneous) motion and so the scaler product $\langle {\bf F},{\bf v}\rangle$, also called the dot product and written ${\bf F} \cdot {\bf v}$, gives the component of the force in the direction of (instantaneous) motion. The instantaneous version of multiplication is integration, and so $W = F\times d$ becomes $$W=\int_{t_1}^{t_2} {\bf F} \cdot {\bf v} \, \operatorname{d}\!t$$

Your paths has the equation $y=x^3$ with $-1 \le x \le 1$. We can parametrise this as $x(t)=t$ and $y(t)=t^3$ where $-1 \le t \le 1$. In vector form: ${\bf x}(t) = t{\bf i} + t^3{\bf j}$. The force can be re-written: \begin{eqnarray*} {\bf F}(x,y) &=& (y^2-x^2){\bf i} + (2-x+y){\bf j} \\ \\ {\bf F}(t) &=& (t^6-t^2){\bf i} + (2-t+t^3){\bf j} \end{eqnarray*} Finally, ${\bf v}(t) = \dot{\bf x}(t) = {\bf i} + 3t^2{\bf j}$. Hence: \begin{eqnarray*} W &=& \int_{t_1}^{t_2} {\bf F} \cdot {\bf v} \, \operatorname{d}\!t \\ \\ &=& \int_{-1}^1 \left((t^6-t^2){\bf i} + (2-t+t^3){\bf j}\right) \cdot \left({\bf i} + 3t^2{\bf j}\right) \, \operatorname{d}\!t \\ \\ &=& \int_{-1}^1 t^6-t^2+6t^2-3t^3+3t^5 \, \operatorname{d}\!t \\ \\ &=& \int_{-1}^1 t^6+5t^2-3t^3+3t^5 \, \operatorname{d}\!t \\ \\ &=& \left[ \tfrac{1}{7}t^7+\tfrac{5}{3}t^3-\tfrac{3}{4}t^4+\tfrac{1}{2}t^6 \right]_{-1}^1 \\ \\ &=& \tfrac{76}{21} \end{eqnarray*}

  • $\begingroup$ A nit: we weren't given units, so don't know that it is joules, ergs, dyne-fathoms or ??? $\endgroup$ – Ross Millikan Oct 15 '13 at 16:27
  • $\begingroup$ I think lb. Angstroms? $\endgroup$ – copper.hat Oct 15 '13 at 16:28
  • $\begingroup$ @RossMillikan One school of thought is that, unless stated, units are SI and so displacement is in metres and force is in newtons, so work must be in joules. Having said that, if this is the case then I ought not mention joules either in order to be consistent with omission of SI units. Thanks Ross. $\endgroup$ – Fly by Night Oct 15 '13 at 16:30

The work done by a force $f$ on a particle moving through a suitably smooth curve $\gamma$ on the interval $[t_0,t_1]$ is given by $W = \int_{t_0}^{t_1} \langle f(t), \dot{\gamma}(t) \rangle dt$.

Here we can parameterize the curve by $\gamma(t) = (t,t^3)$ for $t \in [-1,1]$.

We have $f(t) = F(\gamma(t)) = (t^6-t^2,t^3-t+2)$, and $\dot{\gamma}(t) = (1,3t^2)$.

Hence $ W = \int_{-1}^1 (t^6-t^2+3 t^2(t^3-t+2)) dt = \frac{76}{21}$.


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