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Could someone help me with this Physics question please? Watch

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    I think the value of 'h' is 0.1m from using sin(78)*2. I got 1.9 and I took this away from 2m.
    I don't know how to do the second part, any help would be appreciated.
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    use mgh = 0.5mv^2 and rearrange for v
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    (Original post by DenizS)
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    I think the value of 'h' is 0.1m from using sin(78)*2. I got 1.9 and I took this away from 2m.
    I don't know how to do the second part, any help would be appreciated.
    You are on the right track, but need to be more accurate and round to three decimal places.

    Once you have the height the pendulum falls through, plug that into mgh to get the initial potential energy.

    Them rearrange the kinetic energy formula with v as the subject and using the value of mgh you calculated.
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    (Original post by uberteknik)
    You are on the right track, but need to be more accurate and round to three decimal places.

    Once you have the height the pendulum falls through, plug that into mgh to get the initial potential energy.

    Them rearrange the kinetic energy formula with v as the subject and using the value of mgh you calculated.
    I have rearranged the formula so it is v= the square root of 2mgh/m
    I then substitute in the square root, 2(0.2*9.81*0.1)/0.2. This gives me 1.4.

    Is 1.4 my initial velocity?
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    (Original post by DenizS)
    I have rearranged the formula so it is v= the square root of 2mgh/m
    I then substitute in the square root, 2(0.2*9.81*0.1)/0.2. This gives me 1.4.

    Is 1.4 my initial velocity?
    mgh = \frac{1}{2}mv^{2}

    rearrange this to get v on it's own:

    NB because m is on both sides of the equation it cancels out to leave:

    gh = \frac{1}{2}v^{2}

    now rearrange:

    2gh = v^{2}

    v = \sqrt{2gh}

    Your value of h using trigonometry is not accurate enough. Round this value to 2 decimal places (the question gave g = 9.81ms-2 so you need to be consistent with this.
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    (Original post by uberteknik)
    mgh = \frac{1}{2}mv^{2}

    rearrange this to get v on it's own:

    NB because m is on both sides of the equation it cancels out to leave:

    gh = \frac{1}{2}v^{2}

    now rearrange:

    2gh = v^{2}

    v = \sqrt{2gh}

    Your value of h using trigonometry is not accurate enough. Round this value to 2 decimal places (the question gave g = 9.81ms-2 so you need to be consistent with this.
    Okay, I substituted 0.04 as the change in height into the equation to get 0.7848. Is 0.78 my initial velocity?
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    (Original post by DenizS)
    Okay, I substituted 0.04 as the change in height into the equation to get 0.7848. Is 0.78 my initial velocity?
    You want to check your calculation again.
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    (Original post by DenizS)
    Okay, I substituted 0.04 as the change in height into the equation to get 0.7848. Is 0.78 my initial velocity?
    h = 2 - 2sin(78) = 0.0437 metres

    v = \sqrt{2gh} = \sqrt{2 \rm \ x\ 9.81 \rm \ x\ 0.0437} = ?
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    (Original post by uberteknik)
    h = 2 - 2sin(78) = 0.0437 metres

    v = \sqrt{2gh} = \sqrt{2 \rm \ x\ 9.81 \rm \ x\ 0.0437} = ?
    0.93
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    (Original post by DenizS)
    0.93
    Yes.

    Maximum v is when all of the potential energy is converted to kinetic energy at the bottom of the swing.

    v = 0.93 ms-1 (2 d.p.)
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    (Original post by uberteknik)
    Yes.

    Maximum v is when all of the potential energy is converted to kinetic energy at the bottom of the swing.

    v = 0.93 ms-1 (2 d.p.)
    Thank you
 
 
 
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