0
$\begingroup$

I was given a force, let's say $F = 15xy$ N to the right which moves a system from $A(0,0)$ to $B(3,4)$ and was to find the work done by the force, in Joules.

My approach was by using $\vec{F} = -\vec{\nabla}U$ to find the potential energy in $A(0,0)$ and the potential energy in $B(3,4)$, then using the relationship that $W = -\Delta U$ to find the work done.

Although, I have problems on how to find $U$ based on the equation $\vec{F} = -\vec{\nabla}U$. Based on the problem, I made $\vec{F} = 15 xy \hat{i}$ but I have no luck finding the $U$.

The choices to the problem (in Joules): $540, 405, 1080, 1280, 2700$. I tried another way but the answer was not in the multiple-choice.

Improve this question
$\endgroup$
7
  • $\begingroup$ There is no basis vector attached to force? Or is $ F = xy \vec{i}$...? $\endgroup$
    – Brian
    Commented Oct 19, 2020 at 7:33
  • $\begingroup$ Do you know the direction of $\vec{F}$? I think you should try the integral formula for work. $\endgroup$
    – Guliano
    Commented Oct 19, 2020 at 7:40
  • $\begingroup$ @Buraian The problem only says the force is directed to the right, so I assumed it was in the $\hat{i}$ direction. There was no path given, so I assumed that it's conservative, but I just found out that there's no way for that force to be conservative? $\endgroup$
    – aco
    Commented Oct 19, 2020 at 7:40
  • $\begingroup$ @JulianDeV Yeah, I assumed "right" to be in the direction of $\hat{i}$ but I've got no luck continuing it because the problem doesn't tell me the path the particle took. The previous problems given to me has it but this one doesn't. $\endgroup$
    – aco
    Commented Oct 19, 2020 at 7:44
  • $\begingroup$ I wrote an answer, does it help? $\endgroup$
    – Brian
    Commented Oct 19, 2020 at 7:47

1 Answer 1

Reset to default
1
$\begingroup$

Suppose our force has expression as shown below:

$$ F = F_x \vec{i} + F_y \vec{j} + F_z \vec{k}$$

If a potential function $U$ exists then,

$$ F = -\nabla U = -[\frac{\partial U}{\partial x} \vec{i} + \frac{\partial U}{\partial y} \vec{j}+ \frac{\partial U}{\partial z} \vec{k}]$$

Now, you can compare any one of the coefficients, say we do $x$,

$$- \frac{ \partial U}{\partial x} = F_x$$

We can run a definite integral on both sides to get the potential:

$$ U = - \int F_x dx + h(y,z)$$

We need to include the extraneous $ h(y,z)$ term as all function of that form is differentiated to zero when derivated with $x$. Keeping it in their lets us fit the potential with other force components. So, we must find $h(y,z)$ such that the following system is satisfied:

$$ \begin{bmatrix} \frac{ \partial U}{\partial y} \\ \frac{ \partial U}{\partial z} \end{bmatrix} = - \begin{bmatrix} F_y \\ F_z \end{bmatrix}$$

Also, to check that a potential function exists, just need to check that force field is irrational that is:

$$ \nabla \times F = 0$$

For this case:

$$ ( \frac{ \partial}{\partial x} \vec{i} + \frac{ \partial}{\partial y} \vec{j} + \frac{ \partial}{\partial z} \vec{k}) \times (xy \vec{i}) $$

Doesn't seem this one is conservative 😥 This problem seems to be unsolvable unless you specify a path. Without a lack of context, I'd assume they mean a straight-line path.

On further thought: The force can be made to be conservative if assume that 'right' means the $ \vec{k}$ direction i.e:

$$ F= xy \vec{k}$$

Refer here for more information

Improve this answer
$\endgroup$
1
  • $\begingroup$ I see. I'll stick with the straight-line path one, then, even though the answer is not in the multiple-choice. Thanks! $\endgroup$
    – aco
    Commented Oct 19, 2020 at 7:54

Not the answer you're looking for? Browse other questions tagged or ask your own question.