# m2 work energyWatch

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#1
http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf

q5b

i've been told and we have confirmed in general that the work energy principle is
fs=0.5mv² but how do i apply it to this question?

i have so far

245=0.5x10x11.5² which doesn't help me -.-'
0
3 years ago
#2
(Original post by will'o'wisp)
http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf

q5b

i've been told and we have confirmed in general that the work energy principle is
fs=0.5mv² but how do i apply it to this question?

i have so far

245=0.5x10x11.5² which doesn't help me -.-'
That formula is only true when there is only a change in kinetic energy. But there may be gravitational potential energy change. You could include GPE in the 'work done' part of the formula but from my experience, that confuses students.

It's much better to understand what's going on than remember a formula since the formula could change depending on the question.

In M2 work/energy problems, a particle could lose/gain kinetic energy and lose/gain gravitational potential energy. If there is an overall loss in energy then this loss must be equal to the work done against resistance.

E.g. a particle in travelling from point A to B gains 500J of GPE and loses 700J of KE. So overall the particle has lost 200J of energy.

This energy loss will be equal to the work done against resistance so you would have

200 = Fs

So for this question work through it like this:

Gain/loss in KE =
Gain/loss in GPE =
Work done against friction =

If you work through every work/energy problem like this then you should find them easier.

Post your working if you get stuck.
0
#3
(Original post by notnek)
That formula is only true when there is only a change in kinetic energy. But there may be gravitational potential energy change. You could include GPE in the 'work done' part of the formula but from my experience, that confuses students.

It's much better to understand what's going on than remember a formula since the formula could change depending on the question.

In M2 work/energy problems, a particle could lose/gain kinetic energy and lose/gain gravitational potential energy. If there is an overall loss in energy then this loss must be equal to the work done against resistance.

E.g. a particle in travelling from point A to B gains 500J of GPE and loses 700J of KE. So overall the particle has lost 200J of energy.

This energy loss will be equal to the work done against resistance so you would have

200 = Fs

So for this question work through it like this:

Gain/loss in KE =
Gain/loss in GPE =
Work done against friction =

If you work through every work/energy problem like this then you should find them easier.

Post your working if you get stuck.
so change in energy=fs?

in any case

KE lost=0.5mu²=0.5x10x11.5²=661.2 5
GPE gained=mgh=10gx6.5x 5/13=25g=245
Work done is given in question which is 245

so how do i link these together and gt an answer out?
0
3 years ago
#4
(Original post by will'o'wisp)
so change in energy=fs?

in any case

KE lost=0.5mu²=0.5x10x11.5²=661.2 5
GPE gained=mgh=10gx6.5x 5/13=25g=245
Work done is given in question which is 245

so how do i link these together and gt an answer out?
Your KE lost is wrong. If it was 0.5x10x11.5² then that would mean that the particle has lost all its kinetic energy so it's at rest.

But it may have some speed at point B so could have some kinetic energy at B.

Use to denote the final speed then what is the KE lost?
0
#5
(Original post by notnek)
Your KE lost is wrong. If it was 0.5x10x11.5² then that would mean that the particle has lost all its kinetic energy so it's at rest.

But it may have some speed at point B so could have some kinetic energy at B.

Use to denote the final speed then what is the KE lost?
so then KE lost is 0.5mv² where i don't know the final speed and so 0.5x10xv²=5v²
0
3 years ago
#6
(Original post by will'o'wisp)
so then KE lost is 0.5mv² where i don't know the final speed and so 0.5x10xv²=5v²
When I say 'KE lost' I mean the total KE lost as the particle moves from A to B.

The particle will have more KE at A than it will at B so

KE lost = Initial KE - FInal KE.
0
#7
(Original post by notnek)
When I say 'KE lost' I mean the total KE lost as the particle moves from A to B.

The particle will have more KE at A than it will at B so

KE lost = Initial KE - FInal KE.
oh, i see

so in which case

0.5m(v²-u²)=0.5x10(v²-11.5²)=5v²-661.25
0
3 years ago
#8
(Original post by will'o'wisp)
oh, i see

so in which case

0.5m(v²-u²)=0.5x10(v²-11.5²)=5v²-661.25
That is the change in KE but that is not how much KE is lost.

E.g. the particle has 400J at A and 100J at B so it has lost 400-300=100J.

Or you could say its energy change is 300-400 = -100J. This is what you did.

You could consider change of energy instead of specifically gain/loss but I prefer gain/loss. Let me know if you want to do it the other way.
0
#9
(Original post by notnek)
That is the change in KE but that is not how much KE is lost.

E.g. the particle has 400J at A and 100J at B so it has lost 400-300=100J.

Or you could say its energy change is 300-400 = -100J. This is what you did.

You could consider change of energy instead of specifically gain/loss but I prefer gain/loss. Let me know if you want to do it the other way.
Ok, well seems like i never really understood this topic either ;_______________;

Just talk me through how you'd do this question and i'll work out of the textbook :/

I just don't understand it. Which ever way you think is best for me you teach me.
0
3 years ago
#10
(Original post by will'o'wisp)
Ok, well seems like i never really understood this topic either ;_______________;

Just talk me through how you'd do this question and i'll work out of the textbook :/

I just don't understand it. Which ever way you think is best for me you teach me.
Gain/loss in KE = LOSS

Gain/loss in GPE = GAIN

Work done against friction = 245J

The total loss of energy is equal to the work done against friction.

The total loss of energy will be the loss in KE minus the gain in GPE. Try to think about why this makes sense.

Can you finish the question?
0
#11
(Original post by notnek)
Gain/loss in KE = LOSS

Gain/loss in GPE = GAIN

Work done against friction = 245J

The total loss of energy is equal to the work done against friction.

The total loss of energy will be the loss in KE minus the gain in GPE. Try to think about why this makes sense.

Can you finish the question?
so regardless of gpe and ke

it's always Loss-gain=WD vs Friction
0
3 years ago
#12
(Original post by will'o'wisp)
so regardless of gpe and ke

it's always Loss-gain=WD vs Friction
Yes that's correct.

It could be the KE that is lost and the GPE that's gained or the GPE lost and KE gained.

Also, the resistance may not always be friction.
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#13
(Original post by notnek)
Yes that's correct.

It could be the KE that is lost and the GPE that's gained or the GPE lost and KE gained.

Also, the resistance may not always be friction.
ok thanks guess i'll just kms now then spent about 4 hours on this 1 paper skipped collision bc we haven't learned them and still haven't finished this paper
0
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