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# M2: Power! watch

1. Ok I'm proper sh*tin it about M2 now. I took this book out of the libary this afternoon and i can't get my head around the vector example which is the following:

A constant force F=(-2i+j)N acts on a particle as it moves along a straight from point A, position vector (2i-j)m, to point B, position vector (-i+3j)m in 4 secs. Find the average rate at which F is working?

Some how through some really sh*tty unclear working they get 4 watts?
Go figure!

2. (Original post by Kuz)
Ok I'm proper sh*tin it about M2 now. I took this book out of the libary this afternoon and i can't get my head around the vector example which is the following:

A constant force F=(-2i+j)N acts on a particle as it moves along a straight from point A, position vector (2i-j)m, to point B, position vector (-i+3j)m in 4 secs. Find the average rate at which F is working?

Some how through some really sh*tty unclear working they get 4 watts?
Go figure!

Well power=rate of transfer of energy, so if you move something by a distance d using a force f, the work done is f*d and the power if f*d/t.

In this particular case, find the distance used by pythagoras, multiply by the force and divide by 4.
3. P = Fx/t

|F| = √( 2² + 1²) = √5
|x| = √[(2+1)² + (-1-3)²] = 5

P = 5√5/4 W?

I don't know... I just took the magnitudes of the force and the displacements...
4. i doubt a question as nasty as that would come up. vectors is p3 anyway and although they kinda mix the modules it isnt normally to that extent.
5. Ok I'll write out the example as it is in the book even though I don't understand it. I hope someone can explain it to me in simple terms for me to understand it:

AB(-->) = -rA+rB
= -(2i-j+(-i+3j)
= -3i+4j

work done by F
= F.AB(-->)
= (-2i+j).(-3i+4j)
= 10J

rate of working
= 10J in 2.5 seconds
= 4 watts

Huh!
6. AB is the vector s=(-3i + 4j)

The force is the vector F=(-2i+j)

Work done is force times distance.
In vector notation this is done by doing the dot product of F and s.

F.s = |F||s|.cosø

where ø is the angle between them.
The work done by the force on the particle is the product of that force and the distance moved by that particle. Which means that you have to resolve the force in the direction of AB
It's at some angle or other to AB when applied to the particle.
Doing a dot product operation carries out the angle resolution on the force for you automatically.

F.s = (-2i+j).(-3i + 4j)
F.s = (-2).(-3) + (1).(4)
F.s = 6 + 4
F.s = 10 J

Since 10 J of work are done in 2.5 secs then rate of doing work is 10/2.5

P = 4 Watts
=======
7. (Original post by Fermat)
AB is the vector s=(-3i + 4j)

The force is the vector F=(-2i+j)

Work done is force times distance.
In vector notation this is done by doing the dot product of F and s.

F.s = |F||s|.cosø

where ø is the angle between them.
The work done by the force on the particle is the product of that force and the distance moved by that particle. Which means that you have to resolve the force in the direction of AB
It's at some angle or other to AB when applied to the particle.
Doing a dot product operation carries out the angle resolution on the force for you automatically.

F.s = (-2i+j).(-3i + 4j)
F.s = (-2).(-3) + (1).(4)
F.s = 6 + 4
F.s = 10 J

Since 10 J of work are done in 2.5 secs then rate of doing work is 10/2.5

P = 4 Watts
=======
Wow thats really clear, and I can understand that the only thing that baffles me is that they have come to deduction that the 10J of work are carried out in 2.5 secs when in the question it says 4 secs, is this a typo, or is there some deduction I haven't seen?
8. (Original post by Kuz)
Wow thats really clear, and I can understand that the only thing that baffles me is that they have come to deduction that the 10J of work are carried out in 2.5 secs when in the question it says 4 secs, is this a typo, or is there some deduction I haven't seen?
I was hoping it was a typo, I saw that 4 secs value in your original post and I then worked out P = 2.5 W. I couldn't figure out where I'd gone wrong. Then when you posted the full question and gave the time as 2.5 secs! Well I think it must be a mistake. I caan't see how to get 4 watts over 4 secs!
9. dot.product rule in M2??? its P3 stuff though.... if there any other method?

Edit: Yeah im confused about the 4s to 2.5 sec stuff?
10. (Original post by ResidentEvil)
dot.product rule in M2??? its P3 stuff though.... if there any other method?
thats what i thought. maybe its a different board...i do ocr and there has never been a question like that in all the past papers...however, going by the examiner's magnificent performance displayed in our P3 paper i wouldnt be too surprised.....
11. (Original post by ResidentEvil)
dot.product rule in M2??? its P3 stuff though.... if there any other method?

Edit: Yeah im confused about the 4s to 2.5 sec stuff?
I guess you could resolve the forces in the x direction and then the y direction which is a throw back to M1
12. (Original post by Fermat)
I was hoping it was a typo, I saw that 4 secs value in your original post and I then worked out P = 2.5 W. I couldn't figure out where I'd gone wrong. Then when you posted the full question and gave the time as 2.5 secs! Well I think it must be a mistake. I caan't see how to get 4 watts over 4 secs!
Phew! Thats three of us now that think its a typo, so it must be! I've worried about this question for two hours now and missed the footie and half o big brother. I'm goin now! Thanks for your help!

(Original post by noggin)
thats what i thought. maybe its a different board...i do ocr and there has never been a question like that in all the past papers...however, going by the examiner's magnificent performance displayed in our P3 paper i wouldnt be too surprised.....
Exactly, my maths teacher told me that all the maths papers this summer examinations have been really difficult, he said even the GCSE paper as even "e" popped up. So I'm not ruling anything out!

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