# Relation between conservative force and potential energy

Couple of days before I came across a question something like this

A particle is moving in a conservative force field from point $A$ to point $B$. Let $U_A$ and $U_B$ be the potential energies of particle at points $A$ and $B$ respectively and $W_C$ is the work done in the process in moving the particle from $A$ to $B$. (Take work done to be positive)

And we needed to find the correct relations among some alternatives. What I figured out is as follows:

1. Potential Energy of particle is equal to $negative$ of $work$ $done$ by $conservative$ force.
2. And Work done is equal to Change in Potential energy of particle.

So I found the alternatives as $W_C$=$-(U_B-U_A)$ i.e $W_C$=$(U_A-U_B)$ as correct , but correct answer is $W_C$=$(U_B-U_A)$. I think there is some trick in the last statement of question that "(Take work done to be positive) " But if I take work done to be positive then $U_B$<$U_A$ but this also contradicts the answer as the answer is $U_B$>$U_A$

This made a huge confusion in my mind. Please tell me if I am missing some concept. And do tell me the correct answers too.

• Does my answer makes perfect sense to you..? – Mitchell Feb 23 '17 at 16:26
• @BhavyaSharma Here $W_C$ is work done by conservative force right. $W_C$ should be negative i.e $U_A$-$U_B$ right? But answer is $U_B$-$U_A$. Please explain last loss in energy point. – Piyush Raut Feb 23 '17 at 16:31
• When the body comes down the work done by gravity is positive because the displacement and the Gravitational force are in the same direction. As the body comes down, it gains some kinetic energy because a force (Gravitational force) is acting on it. Work done by conservative force will be -ve when the body goes up and +ve when the body comes down. And when the body goes up (PE)final is greater so change is PE (final-initial) will be +ve and according to the formula we get the work done by conservative force -ve, which is correct.. – Mitchell Feb 23 '17 at 16:36

The work done that we consider here is the work done by external force (or by us).

So if the potential energy at a point is high then it means that the work done by us against the conservative field (say, gravity) will be also be high.

For example, if we lift a body, it's potential energy increases with the height because we are doing work against the conservative force (i.e, gravitational force) and this work (+ve work) gets stored in the body as its potential energy. In other words, higher we lift the body higher will be its potential energy.

Energy conservation plays a very important role here.

Now, if the body comes down the work done by gravity will be positive. The body's potential energy will be converted into kinetic energy. The body is coming from a point of high potential energy to a point of lower potential energy, so this loss will be the gain in the kinetic energy (or the work done by gravity).

Therefore, the work done by the conservative force (gravity) will be equal to the loss in potential energy.

Since the work done by gravity is $+ve$ and there is a loss in potential energy $(\Delta U=-ve)$,

$W_{conservative}=-\Delta U$

The problem is that it is not clear from the question which force is doing the work.
Is $F_{\rm C}$ the external force or the force due to the conservative field?

The change in potential energy of a particle in going from position $A,\, (U_{\rm A})$ to position $B,\, (U_{\rm B})$ is the work done by an external force in moving the particle from $A$ to $B,\,(W_{\rm external, AB})$

$$W_{\rm external, AB}= U_{\rm B} - U_{\rm A}$$

[Liftng a mass $m$ in a gravitaional field strength $g$ a height $h$ you (the provider of the external force) would have to exert an upward force $mg$ and move it through a distance $h$ doing and amount of work equal to $mgh$ which is the change in potential energy.]

The work done in moving the particle from $A$ to $B$ by the force due to the conservative field is

$$W_{\rm field, AB}= -(U_{\rm B} - U_{\rm A})= - W_{\rm external, AB}$$

[The force due to the conservative field $mg$ is downwards and in the opposite direction to the ditection in which the mass is move upward a distance $h$ , so the work done by the force due to the conservative field is $-mgh$ and that by definition is minus the change in potential energy.
So again the change in potential energy is $mgh$.]

• But lets assume that work is done by external force then work being positive we get that $U_B$<$U_A$ again false as the answer is $U_A$<$U_B$. So is solution to the question as in answer book wrong ? Assuming work is done by external force then is the alternative $U_B$<$U_A$ correct ? – Piyush Raut Feb 23 '17 at 15:54
• Work done by external force in lifting a mass from $A$ to $B$ (which is positive as the force that you apply and the direction of movement are the same) you get $U_{\rm B} > U_{\rm A}$ an increase in potential energy of the Earth-mass system. – Farcher Feb 23 '17 at 15:58

Potential Energy of particle is equal to negative of work done by conservative force. And Work done is equal to negative Change in Potential energy of particle. As it is loss of energy from system. So, $$W=-(-(U_b-U_a))$$ $$W=U_b-U_a$$