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19 March, 01:10

A shot-putter moves his arm and the 7.0-kg shot through a distance of 1.0 m, giving the shot a velocity of 10 m/s from rest. Find the average force exerted on the shot during this time.

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  1. 19 March, 04:22
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    F = 350 N

    Explanation:

    Let us consider uniform acceleration,

    m = 7 kg

    Distance moved, s = 1 m

    Final velocity, v = 10 m/s

    Initial velocity, u = 0 m/s

    Find acceleration (Use Newton's third law of motion),

    v^2 = u^2 + 2as

    10^2 = 0 + 2*a*1

    100 = 2a

    a = 50 m/s^2

    Find force,

    F = ma = 7*50

    F = 350 N
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