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28 December, 13:17

1. A truck was pulling a R. V. Away from the end of the cliff. The chain broke and the RV fell to the ground. If the truck had an initial velocity of 2 m/s, what was the distance it fell from the edge of the cliff? The cliff is 280 ft tall. What would be the Force of impact? (R. V=1800 kg)

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  1. 28 December, 13:21
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    The force of impact, F = 17682.18 N

    Explanation:

    Given,

    The initial velocity of the truck and associated velocity of RV, u = 2 m/s

    The height of the cliff, h = 280 ft

    = 85.344 m

    The mass of the RV, m = 1800 Kg

    The total energy of the RV at height, 'h' with velocity, 'v', E = P. E + K. E

    P. E = mgh J

    = 1800 x 9.8 x 85.344

    = 1505468.16 J

    K. E = mv²/2

    = 1800 x 2² / 2

    =3600 J

    Therefore total energy, E = 1509068.16 J

    This is equal to the kinetic energy of RV at the impact.

    The force of impact,

    F = E/h

    = 1509068.16 J / 85.344 m

    = 17682.18 N

    Hence, the force of impact of RV is, F = 17682.18 N
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