Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    8
    ReputationRep:
    For an object has a constant force of 10N applied to it (mass = 1kg). This implies that the acceleration is constant and = 10ms^-2.
    This means that the rate of increase in velocity is constant, hence a velocity displacement graph is a straight line increasing" / " .
    SO If a question as draw the graph of kinetic energy against displacement....

    How come these methods have different outcomes?
    work done = kinetic energy so F X d = KE which gives 10 * 10 = 100
    And so a graph of KE against d would be a straight line "-" form o to 100 J because the area of the force/displacement increases at a constant rate.



    However, if you look at the equation of KE=1/2 mv^2 then that would surely suggest KE is proportional to V^2 so the graph would be a curve as velocity is increasing due to the acceleration?


    e.g. http://slideplayer.com/slide/1671985...aight+line.jpg

    How should the graph look (a straight line 0 to 100 as KE=F*d) or a line curving up form 0 to 100 (As KE is proportional to velocity squared)?
    Offline

    15
    ReputationRep:
    (Original post by splitter2017)
    For an object has a constant force of 10N applied to it (mass = 1kg). This implies that the acceleration is constant and = 10ms^-2.
    This means that the rate of increase in velocity is constant, hence a velocity displacement graph is a straight line increasing" / " .
    SO If a question as draw the graph of kinetic energy against displacement....

    How come these methods have different outcomes?
    work done = kinetic energy so F X d = KE which gives 10 * 10 = 100
    And so a graph of KE against d would be a straight line "-" form o to 100 J because the area of the force/displacement increases at a constant rate.



    However, if you look at the equation of KE=1/2 mv^2 then that would surely suggest KE is proportional to V^2 so the graph would be a curve as velocity is increasing due to the acceleration?


    e.g. http://slideplayer.com/slide/1671985...aight+line.jpg

    How should the graph look (a straight line 0 to 100 as KE=F*d) or a line curving up form 0 to 100 (As KE is proportional to velocity squared)?
    Velocity is not the same as distance moved in direction of force, u seem to be confusing two different situations here.

    Please ask any questions u have so i can help with ur understanding.
    Posted on the TSR App. Download from Apple or Google Play
    • Thread Starter
    Offline

    8
    ReputationRep:
    (Original post by Shaanv)
    Velocity is not the same as distance moved in direction of force, u seem to be confusing two different situations here.

    Please ask any questions u have so i can help with ur understanding.
    OK, I have written out what I am trying to understand with the two methods named scenario 1 and scenario 2. I hope this helps describe what I am stuck with.
    My abbreviation KE represents kinetic energy.

    I am simply trying to sketch how kinetic energy would change over the journey i.e. kinetic energy against displacement.

    Name:  Scan0023.jpg
Views: 17
Size:  480.3 KB
    Offline

    10
    ReputationRep:
    (Original post by splitter2017)
    OK, I have written out what I am trying to understand with the two methods named scenario 1 and scenario 2. I hope this helps describe what I am stuck with.
    My abbreviation KE represents kinetic energy.

    I am simply trying to sketch how kinetic energy would change over the journey i.e. kinetic energy against displacement.

    Name:  Scan0023.jpg
Views: 17
Size:  480.3 KB
    Scenario 2 would be correct. You're also correct in saying that KE = Fd but I don't quite understand how you've got a straight line graph from that equation.

    The only graph you can assume from KE=Fd would be one plotted with force against distance where the area underneath the graph equals KE. And it is not a straight line graph through the origin as the same force is applied constantly over the distance. Name:  image.jpg
Views: 16
Size:  321.3 KB
    Offline

    10
    ReputationRep:
    (Original post by G.Y)
    Scenario 2 would be correct. You're also correct in saying that KE = Fd but I don't quite understand how you've got a straight line graph from that equation.

    The only graph you can assume from KE=Fd would be one plotted with force against distance where the area underneath the graph equals KE. And it is not a straight line graph through the origin as the same force is applied constantly over the distance. Name:  image.jpg
Views: 16
Size:  321.3 KB
    Lol wait. If force is constant and KE=Fd then the gradient of a KE against d graph would have to be constant meaning it cannot be curved.
    Offline

    10
    ReputationRep:
    (Original post by splitter2017)
    OK, I have written out what I am trying to understand with the two methods named scenario 1 and scenario 2. I hope this helps describe what I am stuck with.
    My abbreviation KE represents kinetic energy.

    I am simply trying to sketch how kinetic energy would change over the journey i.e. kinetic energy against displacement.

    Name:  Scan0023.jpg
Views: 17
Size:  480.3 KB
    Maybe the following equation can help:
    v2 = u2 + 2as
 
 
 
Today on TSR

Ask the experts

Advice on your uni application, apprenticeships, getting a job, dropping out of uni and more...

  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Did TEF Bronze Award affect your UCAS choices?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • 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

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