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6 February, 23:03

What average force is needed to accelerate a 9.20-gram pellet from rest to 125 m/s over a distance of 0.800 m along the barrel of a rifle?

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  1. 6 February, 23:25
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    Given:

    u = 0, initial velocity

    v = 125 m/s, final velocity

    s = 0.0800 m, distance traveled

    9.20 g, the mass of the pellet

    If the acceleration is a, then

    0² + 2 * (a m/s²) * (0.800 m) = (125 m/s) ²

    1.6a = 15625

    a = 9765.625 m/s²

    Calculate the force.

    F = (9.20 x 10⁻³ kg) * (9765.625 m/s²) = 98.84 N

    Answer: 98.8 N (nearest tenth)
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