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7 November, 23:49

A 0.650 kg hammer is moving horizontally at 4.00 m/s when it strikes a nail and comes to rest after driving it 1.00 cm into a board. (a) Calculate the duration of the impact.

(b) What was the average force exerted on the nail?

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  1. 8 November, 03:21
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    a) 0.005s

    b) F = - 520 N, this is taking the direction into account.

    Explanation:

    let d be the distance covered by the hummer and nail, a be the acceleration of the hummer and nail

    a) m = 0.650 kg

    Vi = 4.00 m/s

    d = 0.01 m

    (Vf) ^2 = (Vi) ^2 + 2 (a) (d)

    0 = (4) ^2 + 2 (a) (0.01)

    a = - 800 m/s^2

    then:

    Vf = Vi + at

    0 = 4 + (-800) t

    t = 0.005s

    Therefore, the impact lasted for 0.005s.

    b) the average force is given by:

    F = ma = (0.650) (-800) = - 520 N

    Therefore, the magnitude of the average force is 520 N.
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