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    Im stuck on question 5c. Heres the question
    5. A post is driven into the ground by means of a blow from a pile-driver. The pile-driver falls
    from rest from a height of 1.6 m above the top of the post.
    (a) Show that the speed of the pile-driver just before it hits the post is 5.6 m s–1
    . (2 marks)
    The post has mass 6 kg and the pile-driver has mass 78 kg. When the pile-driver hits the top
    of the post, it is assumed that the there is no rebound and that both then move together with
    the same speed.
    (b) Find the speed of the pile-driver and the post immediately after the pile-driver has hit the
    post. (3 marks)
    The post is brought to rest by the action of a resistive force from the ground acting for 0.06 s.
    By modelling this force as constant throughout this time,
    (c) find the magnitude of the resistive force, (4 marks)
    so i got v=5.6 for a
    u=5.2 ms-1 for b
    i've checked the mark scheme and both of the answers are correct
    however for c,
    the mark scheme says
    84 . 5.2 = F . 0.06 – 84g . 0.06
    ⇒ F = 8103.2 N
    which i dont really understand.. is there a specific formula/equation that i need to use for that? from what i understand is that 84x5.2 is probably the magnitude of impulse, and Fx0.06 is because it is acting for 0.06 seconds, but i dont really get is why i have to times 84g by 0.06? and why is the whole thing put in that way?
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    (Original post by tifflcy)
    Im stuck on question 5c. Heres the question
    5. A post is driven into the ground by means of a blow from a pile-driver. The pile-driver falls
    from rest from a height of 1.6 m above the top of the post.
    (a) Show that the speed of the pile-driver just before it hits the post is 5.6 m s–1
    . (2 marks)
    The post has mass 6 kg and the pile-driver has mass 78 kg. When the pile-driver hits the top
    of the post, it is assumed that the there is no rebound and that both then move together with
    the same speed.
    (b) Find the speed of the pile-driver and the post immediately after the pile-driver has hit the
    post. (3 marks)
    The post is brought to rest by the action of a resistive force from the ground acting for 0.06 s.
    By modelling this force as constant throughout this time,
    (c) find the magnitude of the resistive force, (4 marks)
    so i got v=5.6 for a
    u=5.2 ms-1 for b
    i've checked the mark scheme and both of the answers are correct
    however for c,
    the mark scheme says
    84 . 5.2 = F . 0.06 – 84g . 0.06
    ⇒ F = 8103.2 N
    which i dont really understand.. is there a specific formula/equation that i need to use for that? from what i understand is that 84x5.2 is probably the magnitude of impulse, and Fx0.06 is because it is acting for 0.06 seconds, but i dont really get is why i have to times 84g by 0.06? and why is the whole thing put in that way?
    For part c you do suvat to fine your deacceleration, then use F=Ma to find your resistive force if you specify up as positive then resistive force and acceleration is positive giving you, Fr= Ma + weight(mg)
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    (Original post by tifflcy)
    Im stuck on question 5c. Heres the question
    5. A post is driven into the ground by means of a blow from a pile-driver. The pile-driver falls
    from rest from a height of 1.6 m above the top of the post.
    (a) Show that the speed of the pile-driver just before it hits the post is 5.6 m s–1
    . (2 marks)
    The post has mass 6 kg and the pile-driver has mass 78 kg. When the pile-driver hits the top
    of the post, it is assumed that the there is no rebound and that both then move together with
    the same speed.
    (b) Find the speed of the pile-driver and the post immediately after the pile-driver has hit the
    post. (3 marks)
    The post is brought to rest by the action of a resistive force from the ground acting for 0.06 s.
    By modelling this force as constant throughout this time,
    (c) find the magnitude of the resistive force, (4 marks)
    so i got v=5.6 for a
    u=5.2 ms-1 for b
    i've checked the mark scheme and both of the answers are correct
    however for c,
    the mark scheme says
    84 . 5.2 = F . 0.06 – 84g . 0.06
    ⇒ F = 8103.2 N
    which i dont really understand.. is there a specific formula/equation that i need to use for that? from what i understand is that 84x5.2 is probably the magnitude of impulse, and Fx0.06 is because it is acting for 0.06 seconds, but i dont really get is why i have to times 84g by 0.06? and why is the whole thing put in that way?
    Spoiler:
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    In case above wasn't clear.
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    (Original post by Middriver)
    For part c you do suvat to fine your deacceleration, then use F=Ma to find your resistive force if you specify up as positive then resistive force and acceleration is positive giving you, Fr= Ma + weight(mg)
    I used suvat n found a=-260/3 n the markscheme says it got nth to do w suvat
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    (Original post by Middriver)
    Spoiler:
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    In case above wasn't clear.
    ohhhhh alright thank you so much
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    i still dont get the method on the mark scheme tho?
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    (Original post by tifflcy)
    i still dont get the method on the mark scheme tho?
    Is it an official edexcel paper?
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    (Original post by Middriver)
    Is it an official edexcel paper?
    I think so, found it on physicsandmathstutor
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    (Original post by tifflcy)
    i still dont get the method on the mark scheme tho?
    Would be better written as (F-84g)*0.06 = 84 * 5.2

    Then it's just net force * time = impulse.

    The net force is F - 84g since the resistive force acts upwards and the weight 84g acts downwards.
 
 
 
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