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    I'm having an issue with the (c)(i) part of the attached question.

    You will find an attached markscheme with all the answers so you dont have to work out the initial part of the question.

    For the aforementioned part I used the simple idea :

    initial energy = final energy.

    As the cyclists moves up the slope the initial energy is purely kinetic, hence :

    0.5*92*62 = WORK DONE AGANIST FRICTION + 0.5*92*32 +92*9.8*1.3

    Thus, I get the Work done against friction as 70 Joules.

    I am asked to prove that the work done against friction is 670 joules.... What have I missed out ?
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    You have missed out the fact that the cyclist has pedalled up the slope and that the energy he supplied was 600J as calculated in the previous part of the question.
    So the loss in energy against friction is not just that calculated from the difference between the kinetic energy lost and the gain in potential potential energy. The value you have there is what the answer would have been had he not pedalled.
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    (Original post by Stonebridge)
    You have missed out the fact that the cyclist has pedalled up the slope and that the energy he supplied was 600J as calculated in the previous part of the question.
    So the loss in energy against friction is not just that calculated from the difference between the kinetic energy lost and the gain in potential potential energy. The value you have there is what the answer would have been had he not pedalled.
    I dont understand the part in bold.
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    One other thing.

    In projectile motion could it be argued that the vertical component of ACCELERATION is always CONSTANT and acting DOWNWARDS ?

    Ignore air resistance.
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    (Original post by Ari Ben Canaan)
    I dont understand the part in bold.
    Your answer, 70J, would be correct if the cyclist didn't pedal up the slope.
    Then you would just look at the loss in kinetic energy (1240J) and the gain in potential (1170J) to find the energy lost as friction. (70J)
    But as he did pedal, and generated 600J of energy, this energy was also lost against friction bringing the total to 670J.
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    (Original post by Ari Ben Canaan)
    One other thing.

    In projectile motion could it be argued that the vertical component of ACCELERATION is always CONSTANT and acting DOWNWARDS ?

    Ignore air resistance.
    In projectile problems there is only a vertical component of acceleration, which is constant and equal to g. The horizontal motion is not affected by gravity, and the horizontal component of velocity is taken as constant.
    (Yes, these assumptions require there to be no air resistance)
 
 
 
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