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Work Done by a Force 3

https://isaacphysics.org/questions/work_done3?stage=a_level

Could someone explain what the question is saying - I got the answer for all the parts but don't really understand what is going on.
I understand what r is but no idea about F.
r = xi+yj which makes sense to me because the axis for both lines up
but for F...
I initially tried to write F as a column vector so F = (F_0*y/r,-F_0*x/r) - but unsure why y is multiplied by i - shouldn't that be x?

(edited 11 months ago)

Reply 1

mqb2766

Original post by mosaurlodon

https://isaacphysics.org/questions/work_done3?stage=a_level
Could someone explain what the question is saying - I got the answer for all the parts but don't really understand what is going on.
I understand what r is but no idea about F.
r = xi+yj which makes sense to me because the axis for both lines up
but for F...
I initially tried to write F as a column vector so F = (F_0*y/r,-F_0*x/r) - but unsure why y is multiplied by i - shouldn't that be x?

You could think of it in polar form and x=r*cos(theta), y=r*sin(theta). So r would make an angle theta with the x axis and F would be perpendicular as the dot is zero, so it would be acting tangentially to (the point on) the circle which is probably what you want?

If youre unsure, pick an angle like theta=30 degrees, r=1 and sketch it out.

(edited 11 months ago)

Reply 2

mosaurlodon

so something like this?

I still don't really understand how you get original equation for F

Reply 3

mqb2766

Original post by mosaurlodon

so something like this?

I still don't really understand how you get original equation for F

F is given and its always perpendicular to the displacement (x,y), so thats why you have zero work done. It must be something like (y,-x) or (-y,x) to get a zero dot product with (x,y).

(edited 11 months ago)

Reply 4

mosaurlodon

ah I see so you cannot work F out by the diagram itself

the diagram actually makes sense and I guess the reason its flipped to (-y,x) is just so it mathematically works out with the dot product like you said.

Thank you :smile:

(edited 11 months ago)

Reply 5

mqb2766

Original post by mosaurlodon

ah I see so you cannot work F out by the diagram itself
the diagram actually makes sense because it has to go a little left (negative) and up (positive) for it to be perpendicular (for this angle anyway) - and I guess the reason its flipped to (-y,x) is just so it mathematically works out with the dot product like you said.
Thank you :smile:

Actually, Im unsure about the whole question. The F.s formula (which they tell you to use) really refers to the work done in applying a force over the displacement s. Here the motion lies on a circle and the force is acting tangentially to the circle so work will be done along the circumferece as the instantenous displacement is in the same direction. So the work done along that path should be something like
Int F ds = Int F.ds/dt dt = Int F.v dt
and the velocity is parallel to the force.

Either Im completely misunderstanding something or there is a problem with the question.

Edit - as an example, centripetal force which generates circular motion acts along the radial direction (negative) however no work is done as the radial displacement does not change. Its perpendicular to the tangential velocity and all that would be in keeping with Int F.v dt rather than F.r

(edited 11 months ago)

Reply 6

mosaurlodon

So just to like reiterate what you said to check If I understood...

usually with circular motion since the centripetal force acts towards the centre, and not in the direction of the displacement (ie going around the circle) then no work is done,
but with this scenario the force IS acting in the direction so work is done.

and the only way to get that work done would be using an integral rather than a dot product?
i might contact Isaac Physics to see if there is something Im missing.

Reply 7

mqb2766

Original post by mosaurlodon

So just to like reiterate what you said to check If I understood...
usually with circular motion since the centripetal force acts towards the centre, and not in the direction of the displacement (ie going around the circle) then no work is done,
but with this scenario the force IS acting in the direction so work is done.
and the only way to get that work done would be using an integral rather than a dot product?
i might contact Isaac Physics to see if there is something Im missing.

I dropped them a line last night as it seems like an obvious mistake (famous last words) and will let you know what they say. Im surprised that noone has picked it up before, though its easy just to chug through the question without thinking about it and I was guilty of that in the first couple of replies. The question is a maths one (rather than physics) and Id guess it was churned out. The question is incomplete as there is no indication of the length of arc the force acts for which is important as the infinitesimal work done is constant (not zero).

Yes for your basic argument about centripetal / this question.
https://phys.libretexts.org/Bookshelves/University_Physics/University_Physics_(OpenStax)/Book%3A_University_Physics_I_-_Mechanics_Sound_Oscillations_and_Waves_(OpenStax)/07%3A_Work_and_Kinetic_Energy/7.02%3A_Work
is a bit long, but has most of the arguments.

(edited 11 months ago)

Reply 8

mosaurlodon

Thank you so much - lets hope they give us an apology rather than a lecture
(although in my case I wouldnt mind a new lesson)

Reply 9

mqb2766

Original post by mosaurlodon

Thank you so much - lets hope they give us an apology rather than a lecture
(although in my case I wouldnt mind a new lesson)

Had a very brief reply from isaac a day or so ago and their last observation about because the particle does not speed up, then no work is done .... is best left uncommented on. Though I have queried their reply.

However, I looked over the previous two work sheets and they both (correctly) refer to work done as the dot of F with the change in displacment or the (integral of) instantaneous displacement. In particular part c) in sheet 2
https://isaacphysics.org/questions/work_done2?board=1ae25da7-3108-48fc-9a50-2c6acc68a471&stage=all
is very similar to the circular motion set up, although the force is in radial direction (in sheet 3 its tangential) and they correctly make the observation that as the dot of that with the (instantaneous) displacment is zero, then work done is zero. Which is the centripetal force (radial direction) argument, but the opposite of what is done in sheet 3. Ho hum ... Cambridge physics (fill in a suitable phrase here) ...

(edited 10 months ago)

Reply 10

mosaurlodon

I see... from what ive noticed they rarely go into detail - they must get a lot of other questions from a lot of people regarding their challenge questions which would explain the brief comment.
Anyways, thank you for that and all your help and I look forward to seeing what they reply with

Reply 11

mqb2766

Original post by mosaurlodon

Just had a reply where they now accept that both the worksheet and their original reply were wrong. So work is done when the force is in the tangential direction which should have been obvious as you could unroll the circle into a straight line and the constant force would always be heading along that line and this is pretty much textbook definition of (simple) work done.

Reply 12

mosaurlodon

I see.. thank you, so are they going to change the question/previous worksheet?

I also might have time to thoroughly go through the libretext you sent, just so I can confidently understand why/why not work done occurs when the force is either tangential or radial (and to know when to use integration of Int F.v dt)

Reply 13

mqb2766

Original post by mosaurlodon

I see.. thank you, so are they going to change the question/previous worksheet?
I also might have time to thoroughly go through the libretext you sent, just so I can confidently understand why/why not work done occurs when the force is either tangential or radial (and to know when to use integration of Int F.v dt)

The guy said he was looking at adapting the text, though then it would pretty much become question c) on worksheet 2, but its their problem now tbh.

Im not sure Id spend lots of time going through that page (its decent but Im sure there are more concise ones) and thinking of it as
Int F.v dt
is a bit beyond a level.
https://math.colorado.edu/~liuf/activecalc3/Project11_LineIntegralsOverVectorFieldSol.pdf
Not hard, but is based on stuff youve not covered. The key thing is that you want to consider the particles displacement (instantaneous/velocity), rather than the particles position (displacement from the origin) when you do F.s or more properly
Int F.ds
where the dot is for a vector and you integrate between s_1 and s_2. For a constant force this is
F.(s_2-s_1)
hence the dot of force with the displacement, s=s_2-s_1.

(edited 10 months ago)

Reply 14

mqb2766

Original post by mosaurlodon

Looks like our meddling has got them to replace the question
https://isaacphysics.org/questions/work_done3?stage=a_level

Reply 15

mosaurlodon

😂 took them 6 months to do it? well better than my work schedule - maybe they like collate all qs that need editting and do them all at once or smth

"our" meddling - i didnt really do much haha, just sat there wide-eyed and watched the chaos unfold.

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