Work done by gravity on a hanging chain?

[Manish_529]

Homework Statement

A chain of length 'L' and mass 'M' and also has an uniform linear mass density. The chain hangs from a support as one of it's end is attached to it, while the other end is free and is hanging downwards. Find the work done by gravity on the entire hanging chain.

Relevant Equations

W=F.S

I tried taking an element of length dx and tried calculating the force of gravity acting on it so that I could just integrate over the whole chain, but I couldn't figure out what to do of that displacement part since the dx element is not moving as is just at rest (hanging). So, how should I proceed with the integration? (A picture has been attached for reference of the above situation.)

 

[phinds]

If the chain is just hanging there, is any work being done?

 

[Manish_529]

since, there is a difference in the P.E of the 2 ends of the chain, there should be a work done cause the work done by conservative forces (which here is just gravity)= -(change in P.E)

 

[jbriggs444]

since, there is a difference in the P.E of the 2 ends of the chain, there should be a work done cause the work done by conservative forces (which here is just gravity)= -(change in P.E)

Work is done when there is a change in potential energy over time.

A change in potential energy over displacement does not qualify as work and would not have the same units as work.

In an advanced class, you may learn that change in a potential over displacement is called the "gradient" of the potential. For a scalar potential (like gravity) the gradient is a vector. In the case of gravitational potential the gradient is the local acceleration of gravity.

It would be meaningful to ask how much work was done by a chain that was laying motionless on a horizontal frictionless surface and then slowly lowered through a frictionless hole in the ceiling to end up in the pictured configuration.

 

[phinds]

since, there is a difference in the P.E of the 2 ends of the chain, there should be a work done cause the work done by conservative forces (which here is just gravity)= -(change in P.E)

But if it is just hanging there, there IS no change in P.E. --- as jbriggs said, you need a change in P.E. OVER TIME.

 

[Manish_529]

Work is done when there is a change in potential energy over time.

A change in potential energy over displacement does not qualify as work and would not have the same units as work.

In an advanced class, you may learn that change in a potential over displacement is called the "gradient" of the potential. For a scalar potential (like gravity) the gradient is a vector. In the case of gravitational potential the gradient is the local acceleration of gravity.

It would be meaningful to ask how much work was done by a chain that was laying motionless on a horizontal frictionless surface and then slowly lowered through a frictionless hole in the ceiling to end up in the pictured configuration.

Work is done when there is a change in potential energy over time.

A change in potential energy over displacement does not qualify as work and would not have the same units as work.

In an advanced class, you may learn that change in a potential over displacement is called the "gradient" of the potential. For a scalar potential (like gravity) the gradient is a vector. In the case of gravitational potential the gradient is the local acceleration of gravity.

It would be meaningful to ask how much work was done by a chain that was laying motionless on a horizontal frictionless surface and then slowly lowered through a frictionless hole in the ceiling to end up in the pictured configuration.

In case of stationary field like that of gravity the force at a point doesn't depend on the time factor rather it depends on the distance from the body applying it so the force is a function of position then the associated potential energy also must be a function of position meaning the potential energy associated with gravity changes with position and not time.

 

[phinds]

changes with position and not time.

That's quite a trick, changing something's position instantaneously (i.e. taking no time)

 

[Manish_529]

Oh i get it work is done by a force when a body change's it's position with the passage of time which in turn changes the P.E with the position because the body's position was changed. Am I right?

 

[jbriggs444]

Oh i get it work is done by a force when a body change's it's position with the passage of time which in turn changes the P.E with the position because the body's position was changed. Am I right?

Yes. This will do.

Often, the idea of potential energy will be introduced with point-like objects so that this confusion cannot arise. A point-like object cannot be in two places at once. So if there is a change of potential energy, there must be a situation before and a situation after. Some time must have elapsed.

If we allow the possibility of extended objects then we can compute the potential energy for the object by adding up the potential energies for all of the object's parts. We may have to evaluate an integral.

If we want to apply the work energy theorem and equate the change in potential energy with the amount of mechanical work that is performed then we might have to evaluate two integrals and subtract them from each other do determine the change in potential energy.

 

[Steve4Physics]

Oh i get it work is done by a force when a body change's it's position with the passage of time which in turn changes the P.E with the position because the body's position was changed. Am I right?

More or less right but can I add this...

Suppose you have an object weighing 20N. That means the object experiences a gravitational force (its weight) of 20N acting downwards.

If the object moves down by 3m (in the same direction to the force’s direction) the work done by the weight = 20N x 3m = 60J. The change in gravitational potential energy (GPE) is the negative of this, i.e. -60J. So GPE reduces by 60J

If the object moves up by 3m (in the opposite direction to the force’s direction) the work done by the weight = - 20N x 3m = -60J. The change in gravitational potential energy (GPE) is the negative of this, i.e. 60J. So GPE increases by 60J.

But if the object moves horizontally sideways by 3m (perpendicular to the force’s direction, e.g. slides along a horizontal table), the work done by the weight = zero because there is no motion in/against the direction of the weight. The change in GPE is zero.

If the displacement and weight are at some arbitrary angle (e.g. a box sliding along a tilted ramp) a bit of trigonometry is needed.