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# Need help with M2 Work, Energy & Power question watch

1. For part b, why is it KE plus work done against friction = PE lost, rather than KE minus work done against friction = PE lost?

https://ibb.co/cs6eJ8
https://ibb.co/ckvdBT
https://ibb.co/idhmy8
2. (Original post by Mystelle)
For part b, why is it KE plus work done against friction = PE lost, rather than KE minus work done against friction = PE lost?

https://ibb.co/cs6eJ8
https://ibb.co/ckvdBT
https://ibb.co/idhmy8
IF there is any confusion, I find it best to go back to a basic equation:

Energy at start = Energy at end + work done.

I'll use the suffix 1 for the start and 2 for the end. So,

KE1 + GPE1 = KE2 + GPE2 + WD

Rearranging.

WD + (KE2-KE1) = (GPE1-GPE2)

Ie.

Work done + gain in KE = loss in GPE.
3. (Original post by ghostwalker)
IF there is any confusion, I find it best to go back to a basic equation:

Energy at start = Energy at end + work done.

I'll use the suffix 1 for the start and 2 for the end. So,

KE1 + GPE1 = KE2 + GPE2 + WD

Rearranging.

WD + (KE2-KE1) = (GPE1-GPE2)

Ie.

Work done + gain in KE = loss in GPE.
Thanks for helping me out, it makes more sense now! Does work done in your equation mean work out? I usually use the equation KE1 + GPE2 + Work in (e.g. engine of truck) - Work out (e.g. friction) = KE2 + GPE2. Although I think rearranging this would get: Work out + (KE2-KE1) = (GPE1-GPE2) + Work in
4. (Original post by Mystelle)
Thanks for helping me out, it makes more sense now! Does work done in your equation mean work out?
Yes. work done by the system.

I usually use the equation KE1 + GPE2 + Work in (e.g. engine of truck) - Work out (e.g. friction) = KE2 + GPE2. Although I think rearranging this would get: Work out + (KE2-KE1) = (GPE1-GPE2) + Work in
In red above, matches their solution, with "work in " being zero.

I left off "work in" to keep it simple. But in full, my basic equation is.

Initial energy + Work in = Final energy + work out.

Then I can plug in GPE, KE, EPE (when you've covered elasticity) as necessary into each part.
5. (Original post by ghostwalker)
Yes. work done by the system.

In red above, matches their solution, with &amp;amp;amp;quot;work in &amp;amp;amp;quot; being zero.

I left off &amp;amp;amp;quot;work in&amp;amp;amp;quot; to keep it simple. But in full, my basic equation is.

Initial energy + Work in = Final energy + work out.

Then I can plug in GPE, KE, EPE (when you've covered elasticity) as necessary into each part.
Ah I understand now! You’ve explained that really well, thank you ghostwalker!

Elasticity is in M3 right? Won’t be covering that unfortunately, as I only do A Level Maths
6. (Original post by Mystelle)
Ah I understand now!

You’ve explained that really well, thank you ghostwalker!
Elasticity is in M3 right? Won’t be covering that unfortunately, as I only do A Level Maths
You're welcome.

And yep, M3.
7. ghostwalker

Hi again,

Sorry I still can’t quite get my head around the equation: KE gained + Work out = PE lost. I know this can be obtained by rearranging the work-energy principle, but I don’t understand the logic...

It makes sense that, for a given amount of kinetic energy in a particle, the weaker the friction acting against it, the further it travels up a slope before coming to a rest. So KE lost - Work out = PE gained.

But in the equation KE gained + Work done against friction = PE lost, how come increasing the friction causes the particle to lose more PE? Surely friction makes it harder for it to slide down?
8. (Original post by Mystelle)
ghostwalker

Hi again,

Sorry I still can’t quite get my head around the equation: KE gained + Work out = PE lost. I know this can be obtained by rearranging the work-energy principle, but I don’t understand the logic...

It makes sense that, for a given amount of kinetic energy in a particle, the weaker the friction acting against it, the further it travels up a slope before coming to a rest. So KE lost - Work out = PE gained.

But in the equation KE gained + Work done against friction = PE lost, how come increasing the friction causes the particle to lose more PE? Surely friction makes it harder for it to slide down?
A particle travels down a slope and loses 100J of PE. If there was no friction it would gain 100J of KE. But if there is friction then there must be a net loss in energy so the particle e.g. may gain 80J of KE instead and so must lose 20J from working against friction.

KE gained + Work done against friction = PE lost
80J + 20J = 100J

Make sense?

Students can find loss/gain confusing so one other way to do these types of questions is to use the full formula for every question:

KE before + PE before - Work out + Work in = KE after + PE after

"Work out" could be work done against friction and "work in" could be work done from a force pulling the particle.
9. (Original post by Notnek)
A particle travels down a slope and loses 100J of PE. If there was no friction it would gain 100J of KE. But if there is friction then there must be a net loss in energy so the particle e.g. may gain 80J of KE instead and so must lose 20J from working against friction.

KE gained + Work done against friction = PE lost
80J + 20J = 100J

Make sense?

Students can find loss/gain confusing so one other way to do these types of questions is to use the full formula for every question:

KE before + PE before - Work out + Work in = KE after + PE after

"Work out" could be work done against friction and "work in" could be work done from a force pulling the particle.
Hey Notnek! Thanks for breaking this down for me! It all makes perfect sense when I think of the equation as PE lost - Work out = KE gained. So if there was less friction, the particle would gain more kinetic energy as a result. Thank you, you’ve been so helpful!

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