I'm not exactly sure on how to word my question, but, here it goes. I'm having trouble comprehending the definition of work and the relationship it has with force and energy. Mathematically it's simple, and solving problems that involve work is straightforward, however, conceptual I'm having trouble.
If a weightlifter lifts a barbell, the work done is mgh. So why is the work done by the weightlifter equal to gravity? Doesn't the weightlifter have to provide a force greater than the force gravity provides to accelerate, hence the net work should be the net force applied to the barbell?
Yes he does provide a force greater than gravity to make it move but at the top of the lift he does negative work by making the barbell stop (decelerate). It works out so that work done by him is mgh.
Michael Stewart
There are two types of work in this scenario which might make it easier to comprehend.
Positive work: The lifter applies a force to the weight in a direction equal to the displacement. Ie., he applies a force to the weights in the up direction and the weight moves up.
Negative work: This is the work done by gravity. Because gravity applies a force to the weight in the downward direction, this force is opposite to the direction of displacement and is therefore negative.
Mathematically though, once the weight has gone from the ground to the lifted position in the air, the only change in energy that has occurred is equal to the force times distance (mgh), so this is all we consider (Work = change in kinetic energy).
Sebastian Nguyen
>dropping the bar disgusting delet this
Thomas Gomez
Everyone gave retarded answers.
The reason the Work is equal to mgh is because the problem states that he is lifting it with constant velocity. Meaning, net force must be zero.
If there were acceleration involved the problem would look a lot different.
Robert Rivera
You fail soph dynamics. kys
Eli Sullivan
The OP is obviously in his first year of physics. Overly complicated answers aren't going to do anything but confuse him further.
Julian Ramirez
Unlike giving completely wrong answers.
Jordan Price
>I don't know what olympic weightlifting is
Jackson Brown
>Overly complicated answers aren't going to do anything
At least the option to take the red pill is there if OP wants it.
And how is F dot dr overcomplicated?
You find the components of the resistive force, dot it with the displacement, and integrate.
If the resistive force is constant, you don't even need to be able to take an integral. It's just straight up vector arithmetic.
Christian Jones
The "work done" is in regard to an end state as compared to the initial state. The barbell after being lifted compared to on the floor. It doesn't actually measure energy expended by the lifter (like in kWh) but only the amount of potential energy put into the barbell.
Got it?
Hudson Murphy
>olympic lifts >not dropping the bar DYEL detected
Brandon Cruz
>only the amount of potential energy put into the barbell
Not OP. What about the case of cargo being slid along the floor against friction? PE doesn't increase from state 1 to state 2, but work is still done.
Wyatt Rodriguez
Work is being done to overcome friction. Box has the same potential energy before and after.
Work is forceXdistance. Potential energy is mgh. Work is often done to put potential energy into a system, not always.
Benjamin Edwards
>Box has the same potential energy before and after.
Was trying to clarify for OP.
Julian Perez
In this system the initial force is high (higher than the force of gravity) as to get the mass accelerated to a useable velocity. During this initial high force bit the wattage is low (as the object is still slow thus f*m is still not large). Force is decreasing while speed and wattage are increasing until some midpoint where the decrease in force is now more relevant than the increase in speed and the wattage starts to drop. Next the force decreases to be less than that of gravity and thus the speed begins to drop. At this point the momentum of the object is being converted to potential energy and (assuming a perfect efficiency system) if you were to measure this shift from lifting wattage to momentum expending wattage you would find that the extra energy applied at the start went here
Christopher Sanders
Work has nothing to do with the force expended. It's all about the force you're working against.
The force you're working against is the object's mass times the acceleration due to gravity. Then you take the resultant (opposite) of that and dot it with the displacement. In this case, the displacement is directly opposite the resisting force, so you just multiply it times the distance (h) from the datum.
Christian James
OP here, thanks for all the answers.
I seem to have found out why I'm so confused. For instant, what is the work that the elevator must provide to lift passengers? The answer is mgh.
I guess if there is an acceleration, that would be explicitly mentioned, otherwise velocity is assumed to be constant? Hence the work that the elevator provides is the negative of mgh.
Is it also implied that the work done by the elevator is the moment the elevator is already in motion? Does it disregard the initial moment the elevator accelerates from rest?
So if a question explcitly asks what is the work required to lift an elevator from rest, the answer would be >mgh?
Parker Hall
> at the top of the lift he does negative work by making the barbell stop (decelerate).
Yeah, I remember all the times I was lifting weights and I forgot to decelerate the weight and I was lifted off the ground.
Fucking hell this board is retarded sometimes.
Benjamin Walker
You shouldn't refer to it as work. Work is forceXdistance. Different than potential energy (mgh).
Gavin Lewis
He's not wrong, if you were to lift a weight and throw it upwards and hold on it has the potential energy (literally) to do the work on you now and lift you off the ground. His explanation is retarded because gravity slows down the weight since you're lifting against it and the instant you stop applying a force on the weight it will accelerate towards the ground but desu maybe you should lift less and study more if you don't understand that simple concept.
Jeremiah Howard
He does not do negative work. That is wrong. All the work he does is in the positive (upwards) direction.
Colton Flores
Where can I learn about the F*dr integral. I don't know why but it just hasn't clicked for me.
Jason Barnes
>Doesn't the weightlifter have to provide a force greater than the force gravity provides to accelerate That's why it has speed. Just look at where the bar stops to find out the work done.
Jack Lewis
The lifter does not do "negative work", but the gravity force does. The work done to accelerate the bar with an upwards force at the start of the lift is equal and opposite to the work done to accelerate the bar in the downwards direction by the force of gravity at the end of the lift. I guess this cancels.
Leo Wilson
You are literally trying to make your retardation contagious. You need to quarantine that shit, bro.
Anthony Hall
Pretty much any intro-level college physics text, or that link I posted back up the thead a ways.
Joshua Diaz
OP, that work is a nonconservative force that is converted into Gibbs free energy. The total sum of all calculated energies will equal 0 and the quantity called the Lagrangian which is potential minus Gibbs is always less than 0.
