a 5kg bag falls a verticle height of 10m before hitting the ground. Ek=1/2mv^2 and Ep=m g h (assume g=9.8N/kg). Using these two

Question

a 5kg bag falls a verticle height of 10m before hitting the ground.
Ek=1/2mv^2 and Ep=m g h (assume g=9.8N/kg).
Using these two equations, calculate the speed the bag reaches before hitting the ground.

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Eirian 2 months 2021-07-23T12:51:26+00:00 1 Answers 4 views 0

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    2021-07-23T12:53:21+00:00

    Answer:

    u = 7m {s}^{ - 1}

    Explanation:

    We know that when we don’t have air friction on a free fall the mechanical energy (I will symbololize it with ME) is equal everywhere. So we have:

    me(1) = me(2)

    where me(1) is mechanical energy while on h=10m

    and me(2) is mechanical energy while on the ground

    Ek(1) + DynamicE(1) = Ek(2) + DynamicE(2)

    Ek(1) is equal to zero since an object that has reached its max height has a speed equal to zero.

    DynamicE(2) is equal to zero since it’s touching the ground

    Using that info we have

    m \times g \times h   =   \frac{1}{2}  \times m \times u {}^{2} \\

    we divide both sides of the equation with mass to make the math easier.

    9.8 \times 10 =  \frac{1}{2}  \times u {}^{2}  \\  \frac{98}{2}  = u {}^{2}  \\ u { }^{2} = 49 \\ u = 7

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