Why is change in potential energy is always negative?

[jackyyoyoyo]

I'm studying relation between conservative force and potential energy,and getting a big question on change in potential energy is always negative.

For gravitational PE ,when an object is lifting up, it's work done is negative(opposite direction). so the change in work done is negative. On the other hand, when an object is falling down, it's work done is positive(same direction). And the change in work done is also positive.
Why?
Am i wrong in change in work done(final-initial)?
Thanks in advance.

 

[gleem]

When an object is lifted the lifting force which is opposing gravity is upward in the same direction as increasing height therefore the work done on the object is positive. Also as the object is lifted its potential energy is increase so the difference is positive. When an object falls due to the force of gravity the work is done by gravity on the object is in the direction of decreasing height and therefore the work is negative. As the object fall its potential energy decreases.

 

[drvrm]

I'm studying relation between conservative force and potential energy,and getting a big question on change in potential energy is always negative.

i think you are asking 'why force is defined as negative rate of change of potential'?
In attractive force field the potential is zero at infinity and as a body moves towards the center of the force , the body goes in bound state and bound state energy is negative - a positive energy means a free state from the field of force .

imagine an electron bound to the nucleus -the energy of electron will be negative till it is bound .
now suppose you are bringing unit mass in a gravitational field of say Earth from infinity to a point at radial distance r and calculate the potential at r...the potential is a negative quantity and the amount of work done was by the field rather than any external agency.

if you try to take a body from r to infinity then the external agency has to do work in reaching zero potential energy.
If you take repulsive field then the work done in bringing a body being repulsed by center of force from infinity to a point r in the field will be done by external agency.

potential energy is scalar quantity and a change in scalar can be either positive or negative but force is a vector and the negative rate of change of potential provides you force -its magnitude as well as its direction.

 

[lychette]

gravitational potential energy is taken to be zero at infinity so all PE must be negative!
Raising a mass increases its PE so it becomes less negative...but it is still a negative quantity.

 

[vanhees71]

I'm not sure whether I understand you question right, but it's just by definition that if the force is conservative the sign of the potential is such that
$$\vec{F}=-\vec{\nabla} V.$$
For a point particle close to Earth the gravitational force is constant, $F=m \vec{g}$, and thus the potential
$$V=-m \vec{g} \cdot \vec{x}.$$
If you define the $z$ direction of your Cartesian coordinate system to point "up", i.e., against $\vec{g}$, then you have $\vec{g}=-g \vec{e}_z$ and thus
$$V=m g z.$$
So the potential energy becomes larger when you lift the particle "up" and smaller when you let it fall "down".

The total energy is constant (because $V$ is not explicitly time dependent), i.e.,
$$E=\frac{m}{2} \dot{\vec{x}}^2+V(\vec{x})=\text{const}.$$

 

[lychette]

I'm not sure whether I understand you question right, but it's just by definition that if the force is conservative the sign of the potential is such that
$$\vec{F}=-\vec{\nabla} V.$$
For a point particle close to Earth the gravitational force is constant, $F=m \vec{g}$, and thus the potential
$$V=-m \vec{g} \cdot \vec{x}.$$
If you define the $z$ direction of your Cartesian coordinate system to point "up", i.e., against $\vec{g}$, then you have $\vec{g}=-g \vec{e}_z$ and thus
$$V=m g z.$$
So the potential energy becomes larger when you lift the particle "up" and smaller when you let it fall "down".

The total energy is constant (because $V$ is not explicitly time dependent), i.e.,
$$E=\frac{m}{2} \dot{\vec{x}}^2+V(\vec{x})=\text{const}.$$

A 'larger' potential energy means 'less' negative.

 

[jackyyoyoyo]

Thanks for response
Now I can figure out why I'm trapped

 

[robphy]

As @vanhees71 says, the minus sign is a definition. With the sign choice as "minus the gradient", the work done by conservative forces can be brought to the kinetic-energy side of the work-energy theorem so that one can state that the net work done by nonconservative forces is equal to the total energy (kinetic plus potential).

Without the minus sign in the definition of potential energy, one would not be able to use the word "total" (implying sum)... instead one would need "kinetic energy minus the (alternately defined) potential energy".