Hi there all: I have a problem!
I need to find the work done on a particle that moves from $(0,0)$ to a point $(1,1)$ by a strait line $y=x$. The force acting upon the particle is $F = (y , 2x$).
Now how i have attempted the problem is I paramaterized the line $x = y$ to be $r = < t, t >$ by letting $x = t$ which means that $y = t$ too. (I'm hoping that this is how you do it, it's been a while since ive done that maths course)
I know the formula for work is: The integral of $F\cdot dr$ – thus meaning I have to differentiate $r$ to get $dr = < 1, 1>$ then finding the force acting on the particle using $F$ and the particles position of $(x,y)$ to be $(1,1)$ (i.e. i substituted this point into $F$) so $F= (1,2)$.
Then taking the dot product between the two i get $3$. Then taking the integral between $1$ and $0$ I get the answer to be $3$. Now the answer in the textbook says the answer is $1.5$ 🙁 bummer.
Where have I gone wrong? Do i need to paramiterise F differently? or.. i dno ??? :S
Best Answer
In a different notation (see below) and
without making the parametrizationusing $x$ as the parameter.The work $W$ carried out by the force $\overrightarrow{F}= y\overrightarrow{i}+2x\overrightarrow{j}$ acting on the particle, when it moves from $P(0,0)$ to $Q(1,1)$ along the line $\gamma :y=x$ (see picture), is given by the integral
$$\begin{eqnarray*} W &=&\int_{\gamma }\overrightarrow{F}\cdot \overrightarrow{dr}\qquad \gamma :y=x,\text{ from }P(0,0)\text{ to}\;Q(1,1) \\ &=&\int_{\gamma } \left( y\overrightarrow{i}+2x\overrightarrow{j} \right) \cdot \left( dx\overrightarrow{i}+dy\overrightarrow{j}\right) \\ &=&\int_{\gamma }y\text{ }dx+2x\text{ }dy,\qquad y=x,\; \; dy=dx \\ &=&\int_{0}^{1}3x\text{ }dx \\ &=&3\int_{0}^{1}x\text{ }dx=3\times \left. \frac{x^{2}}{2}\right\vert _{0}^{1}=3\times \frac{1}{2}=1.5. \end{eqnarray*}$$
Notation: