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    Can someone help me answer q5 and q7 of this exam paper.. I know they maybe basic but i am revising from it so please help!! Please can someone explain with the answer if possible

    Questions in attachment!

    Thanks in advance
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    7 part a: is area under the graph between 0 and 4.

    7 part b: If velocity is not increasing or decreasing it is constant which must make the acceleration ....
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    (Original post by stjohnpeters)
    7 part a: is area under the graph between 0 and 4.

    7 part b: If velocity is not increasing or decreasing it is constant which must make the acceleration ....
    Dont really understand how to to da 7a but for b i know its zero. What do you say about q5?? can you explain with answer for question 7?
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    area under the graph is a triangle... 1/2 x base x height


    then do the other part which is a square base x height
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    5 part b

    energy stored is = workdone...

    Wd= F x S

    displacement = 0.06
    force = 9

    0.06 x 9 = 0.54
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    (Original post by stjohnpeters)
    5 part b

    energy stored is = workdone...

    Wd= F x S

    displacement = 0.06
    force = 9

    0.06 x 9 = 0.54
    Thats what i thought the answer would be? but the mark scheme says its 0.27J?? do you know why that is??
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    **** knows :confused:
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    Only thing i can think of is that considering that it extension and force

    work done = area under the graph...

    1/2 x 0.54 = 0.27j
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    (Original post by stjohnpeters)
    Only thing i can think of is that considering that it extension and force

    work done = area under the graph...

    1/2 x 0.54 = 0.27j
    Ahh yes your right!!!

    Thanks alot man!!

    Will get back to you on q8 to see if your method of area works!
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    (Original post by mujahid_e3)
    Can someone help me answer q5 and q7 of this exam paper.. I know they maybe basic but i am revising from it so please help!! Please can someone explain with the answer if possible

    Questions in attachment!

    Thanks in advance
    q5: Stiffness = Force/Extension = Gradient of graph.

    Gradient = 9/0.06(remember that the y-axis is in cm and the question wants Nm^-1) = 150 Nm^-1

    Energy Stored = Area under graph = 0.5x0.06x9 = 0.27J

    q7: v = s/t => s = vt = area under graph = (0.5x20x2)+(2x20) = 60m

    Acceleration at 3s is 0 ms^-2 because the gradient represents acceleration, and the gradient at 3s is 0. (The velocity isn't changing.)
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    (Original post by Unkempt_One)
    q5: Stiffness = Force/Extension = Gradient of graph.

    Gradient = 9/0.06(remember that the y-axis is in cm and the question wants Nm^-1) = 150 Nm^-1

    Energy Stored = Area under graph = 0.5x0.06x9 = 0.27J

    q7: v = s/t => s = vt = area under graph = (0.5x20x2)+(2x20) = 60m

    Acceleration at 3s is 0 ms^-2 because the gradient represents acceleration, and the gradient at 3s is 0. (The velocity isn't changing.)
    You my friend are a genius??? Thanks alot man!!
    You helped me alot!!

    THANK YOU
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    (Original post by mujahid_e3)
    Thats what i thought the answer would be? but the mark scheme says its 0.27J?? do you know why that is??
    Work Done = Fs, and in this case work done=elastic potential energy gained.

    However, because F has changed over time, you need to use the AVERAGE force in the equation. In this case, average force = (9+0)/2 = 4.5N. Wd = Fs = 4.5x0.06=0.27.

    So using the equation for Work done works, but you need to be careful about what you use as 'F' in the equation. btw, I know you can get perhaps get a faster response here, but this is the forum you should go to for Physics questions: http://www.thestudentroom.co.uk/forumdisplay.php?f=131
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    (Original post by mujahid_e3)
    You my friend are a genius??? Thanks alot man!!
    You helped me alot!!

    THANK YOU
    You're welcome.
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    (Original post by Unkempt_One)
    Work Done = Fs, and in this case work done=elastic potential energy gained.

    However, because F has changed over time, you need to use the AVERAGE force in the equation. In this case, average force = (9+0)/2 = 4.5N. Wd = Fs = 4.5x0.06=0.27.

    So using the equation for Work done works, but you need to be careful about what you use as 'F' in the equation. btw, I know you can get perhaps get a faster response here, but this is the forum you should go to for Physics questions: http://www.thestudentroom.co.uk/forumdisplay.php?f=131
    Thanks will do!
 
 
 
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