Question

Imagine this...

An object is doing work about (200J). A strong gravitational force is applied on that object... Would its total force in this model W = F x D be altered?

So lets say in "general" an object is already doing work and force is applied on it will it increase the work done in a system or will it decrease it?

I had this debat a few days ago with a friend about this topic never knew how to counter it. He kept saying that if more force is applied on an object that is doing work it could help it achieve that work faster/greater/ then usual. I said well maybe...

Answer 1

Force is a vector, not a scalar, so just throwing in some force in some random direction does not increase the work done on an object.

Answer 2

for instance a ball that is thrown (has had a force exerted on it) has the drag of air resistance acting against itbecause the force of air resistance acts opposite that of the force accelerating the ball, the total work done on the ball is less than it would be without the "extra" force if drag.

Answer 3

Same as friction too! Its acted on an object in motion( air resistance is kinda similar).

Answer 4

exactly :)

Answer 5

one formula for the work done on an object is simply the change in energy of that object, kinetic plus potential
imagine the ball moving perfectly horizontally. we would find that in an environment with no air or gravity the final work done on energy of the ball is proportional only to its change in velocity (kinetic energy)\[W\propto\Delta K=\frac12m(\Delta v^2)\]so if the object is moving slower due to air resistance, friction, gravity, whatever, then the total work done on the object is less than it would be without that resisting force.

Answer 6

But this actually got me into thinking... If a force was "directed" on an object could accelerate it and increase its work wouldn't it? Like friction imagine it as a ( - ) effect. Lets say we were able to use a force in a ( + ) effect on a object do you see were I'm going with this? (THIS REEEEALY got me to think of a lot of deep weird complications)

Answer 7

"so if the object is moving slower due to air resistance, friction, gravity, whatever, then the total work done on the object is less than it would be without that resisting force." See force is kinda altering the work process of that ball isn't it? Because it a resistance greater then the work itself.

Answer 8

basically, "yes" to your post that is two up
as far as your last post I'm not sure what you mean by "the work process is different" and that stuff about the resistance being greater than the work.

again note a major difference between vector and scalars: you cannot compare resistance (a force, which is a vector) with work (which is a change in energy, which is a scalar)

A major problem in explaining this thing is that your formula for work is incomplete in many ways. the actual formula for the work done on an object is\[W=\int_a^b\vec F\cdot d\vec s\]
I could explain some details, but I don't know how much you know of vectors and calculus

Answer 9

Well I'm pretty bad at calculus. I'm taking a few classes to improve right not so if you could explain things more interms of a general physics that would be great!

I do know that a vector has both magnitude & direction and a scalor lacks direction but has a magnitude.

What I ment with "work process" is just plain old simple work.

So you do agree that "Force" if directed in a certain direction could help a system/object that is doing work? I kinda felt that would make sense. Sinse the Work model with always involve force so if more force is applied maybe more work is. If force is applied AGAINST an object less work is done because of the resistance.

Now its making more sense and building my final answer to my friend :)

Answer 10

Sorry for my bad grammer,spelling... I get this rush mode that kinda handicaps my typing lol.

Answer 11

Yes, a force acting in the direction of motion (or has some component acting in the direction of motion) will increase the work being done on the object.
If the components of the additional force do not act along the path of motion of the object, then no extra work is done on the object.

One concept would make this much easier to explain: do you know what a dot product is?

Answer 12

"the dot product or scalar product is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number obtained by multiplying corresponding entries and then summing those products. " Makes A LOT of sense! Thank you!!

Answer 13

btw if we look at the situation of the ball and add in the force of gravitynotice that the force of gravity does not act along the line of motion of the object; it is perpendicular to it. That means that it neither does work on or against the object, because no components of the gravity force vector are either parallel or anti-parallel to the direction of motion of the object.

Answer 14

if you understand the dot product this should make sense, since the dot product of two perpendicular vectors is zero, and as I said the real formula for work is a vector dot product, not a regular multiplication\[W=\int_a^b\vec F\cdot d\vec s\] ^
dot product

Answer 15

welcome!

Answer 16

But if "theocratically" gravity went alone perpendicular along the lines of motion then would increase! Many many thanks again :)

Answer 17

exactly, for a falling object gravity is doing work on the object (making the total work done on the object more). for an object moving upwards gravity is doing work against the object (making the total work done on the object less).
welcome again!

Answer 18

This is really interesting... I come back reading this question and would understand many things in our daily lives following this principle! Car,planes,trains,ships,motors,pumps,etc... They all follow the exact same rule which is: W=∫baF⃗ ⋅ds⃗

Sometimes life may seem all black in white until you ask the question and all of a sudden its showing its true colors!

Answer 19

:)

Answer 20

However, Lorentz Force does not follow this concept.

Force is perpendicular. Still work is being done... @TuringTest could you explain more? Because we agreed in order for work to be done or if force to increase work on an object it has to be parallel with its motion. "Lorentz Force".

Answer 21

In which situation involving the Lorentz force are you talking about?

If you mean a charged particle that is in an electric field, it will move along the field lines, so work is being done on the particle.
If you mean a situation like a charged particle that is in a magnetic field, then the particle will just travel in circles (or a helix depending on the angle of the velocity of the particle relative to that of the magnetic field), with the magnetic force always pointing towards the center of the circle(the X's signify that the magnetic field is going into the page)

The force on the particle due to the magnetic field \(\vec B\) is given by\[\vec F=q\vec v\times\vec B\]that \(\times\) in the middle is a vector cross-product (not sure if you know what that is yet...) which means that the force is always perpendicular to \(both\) the velocity of the particle and the direction of the B-field (hence no change in kinetic energy, so no work being done in that department). If the B-field constant it does not change the velocity of the particle, since no component of the magnetic force acts along the line of motion of the particle. Notice also that this force causes the particle to move in a circle, which is a \(closed~~path\) , so the net displacement is zero, hence no work is being done for that reason as well (since the change in potential energy between any two points in the path of the particle is zero).
So for the particle in the constant magnetic field, the Lorentz force does no work because the magnetic force is always perpendicular to the direction of motion, and the displacement along the direction of that force is zero. So there are two reasons that a constant B-field does no work on a charged particle.
This is true of constant gravitational and electrical fields as well; the closed path line integral of any "conservative vector field" is zero (the concept of a conservative vector field is most likely beyond your understanding of calculus at this point, but basically it means a constant force field in physics.... at least for the purposes of this discussion). If I push a block of wood in a circle, starting at rest and ending at rest, even though I feel like I'm pushing on it the net force being done is zero.Why?

1) The net displacement is zero, so the displacement part \(d\vec s\) of \(\int\vec F\cdot d\vec s\) is zero, so the integral is zero.
2) Work is change in kinetic energy plus change in potential energy. It started and ended at rest, so the change in KE=0 and it came back to the same location it was before, so change in U=0 as well, hence the total work is zero.
3) For every bit of work I do on the block, friction is doing work \(against\) the block. Unless I overcome friction and make the block move faster by the end of the journey than before (i.e. increase its kinetic energy) then the forces exerted by friction and I are acting in equal and opposite directions, and so cancel.

Now, if the electric or magnetic fields are changing, they are not conservative, and that means that work can get done on the particle; but only in a non-conservative field.

In short, the magnetic Lorentz force does not necessarily do work on a charged particle unless the B-field is changing (which induces an electric field that \(can\) do work on the particle)

The electric Lorentz force can do work on or against the particle, but only along the direction of the particles motion.

Many of these concepts will become more clear when you further study vector calculus.

If I have failed to explain anything to your satisfaction then I suggest asking @Jemurray3 or @JamesJ to clarify as they are far more knowledgeable than I. In fact, it would be nice if they could tell me whether or not I have explained it properly.