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21 December, 12:33

A force of 1.200*103 N pushes a man on a bicycle forward. Air resistance pushes against him with a force of 615 N. If he starts from rest and is on a level road, what speed? will he be going after 25.0 m?

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  1. 21 December, 14:03
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    15.45 m/s

    Explanation:

    F = 1.2 x 10^3 N, Fk = 615 N, u = 0, s = 25 m

    Let the speed after traveling 25 m is v.

    mass = F / g = 1200 / 9.8 = 122.45 kg

    Net force, Fnet = 1.2 x 1000 - 615 = 1200 - 615 = 585 N

    acceleration = Fnet / mass = 585 / 122.45 = 4.78 m/s^2

    Use third equation of motion

    v^2 = u^2 + 2 a s

    v^2 = 0 + 2 x 4.78 x 25 = 238.88

    v = 15.45 m/s
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