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    Okay I really need help with this one and it would be so nice if someone could explain it to me.

    Imagine we have two springs and they are both stretched by the same force.
    The spring constant of the first spring is twice the spring constant of the second spring.
    If the energy stored in the second spring is E, what's the energy stored in spring one?
    a) E/4
    b) E/2
    c) E
    d)2E
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    You will already know F=kx

    The energy can be thought of either as the area under the F vs X graph, or the average force x the total extension - both give E = 1/2 Fx

    Often, we subsitute the first into the second to give E = 1/2 (kx) x = 1/2 kx².

    That's not really much use to you here becauese for the same F, the extension wll also change because of k, So what I'd do is change the first equation to x = F/k, substitute that into the second one instead. Then you'll have an expression for E in terms of only F and k and you should be able to answer the question.

    Try it, if you get stuck come back and ask again.
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    It will be d) 2E

    energy is the area under force extension graph which is a straight line graph given by F=kx where k is the gradient of the graph

    if the gradient is twice as much then the area under the graph is twice as much, as its a simple linear relationship


    example integrate y=x between 0 and 5. answer = 12.5

    integrate y=2x between 0 and 5. answer is 25 exactly double
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    (Original post by phys981)
    You will already know F=kx

    The energy can be thought of either as the area under the F vs X graph, or the average force x the total extension - both give E = 1/2 Fx

    Often, we subsitute the first into the second to give E = 1/2 (kx) x = 1/2 kx².

    That's not really much use to you here becauese for the same F, the extension wll also change because of k, So what I'd do is change the first equation to x = F/k, substitute that into the second one instead. Then you'll have an expression for E in terms of only F and k and you should be able to answer the question.

    Try it, if you get stuck come back and ask again.
    Thank you so much! this was very helpful
 
 
 
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