Turn on thread page Beta
    • Community Assistant
    • Thread Starter
    Offline

    19
    Community Assistant
    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
    • Study Helper
    Offline

    15
    Study Helper
    (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.
    • Community Assistant
    • Thread Starter
    Offline

    19
    Community Assistant
    (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
    • Study Helper
    Offline

    15
    Study Helper
    (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.
    • Community Assistant
    • Thread Starter
    Offline

    19
    Community Assistant
    (Original post by ghostwalker)
    Yes. work done by the system.



    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.
    Ah I understand now! You’ve explained that really well, thank you ghostwalker! :yy:

    Elasticity is in M3 right? Won’t be covering that unfortunately, as I only do A Level Maths
    • Study Helper
    Offline

    15
    Study Helper
    (Original post by Mystelle)
    Ah I understand now!
    :cool:

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

    And yep, M3.
    • Community Assistant
    • Thread Starter
    Offline

    19
    Community Assistant
    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?
    • Community Assistant
    • Study Helper
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    (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.
    • Community Assistant
    • Thread Starter
    Offline

    19
    Community Assistant
    (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! :cute:
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: May 25, 2018
The home of Results and Clearing

1,790

people online now

1,567,000

students helped last year

University open days

  1. London Metropolitan University
    Undergraduate Open Day Undergraduate
    Sat, 18 Aug '18
  2. Edge Hill University
    All Faculties Undergraduate
    Sat, 18 Aug '18
  3. Bournemouth University
    Clearing Open Day Undergraduate
    Sat, 18 Aug '18
Poll
A-level students - how do you feel about your results?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.