# Work done by a conservative force

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1. Aug 22, 2015

### Titan97

1. The problem statement, all variables and given/known data
The potential energy $U$ of a particle of mass 1kg moving in x-y plane obeys the law $U=3x+4y$. x and y are in meters. If the particle is at rest at (6,8) at time t=0, then find the work done by conservative force on the particle from initial position to the instant when it crosses the x-axis.

2. Relevant equations
$\vec F=-\frac{\partial U}{\partial x}-\frac{\partial U}{\partial y}-\frac{\partial U}{\partial z}$
(This equation is not taken from any book. I thought the relation between F and U was just like the relation between electric field and potential)

3. The attempt at a solution

Using the equation: $F=-3\hat i-4\hat j$
Since no other forces are acting, the particle will move in the direction of acceleration. I also have to find the x-coordinate when it crosses x-axis. Acceleration is at an angle $\tan^{-1}\big(\frac{4}{3}\big)$ with the horizontal towards the 3rd quadrant. Hence the particle moves along the line $y=\frac{4}{3}(x-6)+8$.
So the x intercept is 0. Hence the total distance moved by the particle is 10m. And work done is 50J.
Is this correct?

2. Aug 22, 2015

### BvU

It is. A lot of work done by you. The work done by the field can also be compared to the difference U(6,8) - U(0,0). Could that be a concidence or is there more to that ?

3. Aug 22, 2015

### Titan97

Work done by conservative force is path independent. But I still need to do "work" to find the final position . Or is there another way to solve the problem without having to find the final position?

4. Aug 22, 2015

### BvU

Force always points at (0,0) !

5. Aug 22, 2015

### Titan97

Why should it always point at (0,0)? Is it because the force is conservative?

6. Aug 22, 2015

### BvU

No, but starting at (6,8) there is no component other than in the direction of (0,0)

7. Aug 22, 2015

### Titan97

I dont understand that. Is it because the minimum magnitude of potential energy is at (0,0)? Since a particle tends to reach minimum potential energy. If thats the case, then is it true for $U=2x^2+1$? Since force points at (0,1).

8. Aug 22, 2015

### BvU

Potential energy is a scalar. It has a value. Vectors have a magnitude. Unfortunately daily language mixes them up.

Lines of constant U are straight lines for $U=3x+4y$. Constant U means no force component along that line. The force points perpendicular to those equipotential lines, so once on a line through the origin means following a straight path through the origin if starting from x > 0 and away from the origin when starting from x $\le$ 0.

For $U=x^2+1$ lines of constant U are straight lines also. The force always points towards the y axis, not at 0,1 !