HTD asked in Science & MathematicsPhysics · 1 month ago

A 500 kg car slows down due to friction from 30 m/s to 20 m/s. What was the work done against friction. ?

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  • 1 month ago

    I don't know what you mean by "work done against friction".

    The net work done on the car will equal the car's change in kinetic energy.

    Kinetic energy:   KE = 0.5 mv^2

    The change in kinetic energy of the car is:0.5*500kg*(20m/s)^2 - 0.5*500kg*(30m/s)^2 = -125,000J

    If friction is the only force slowing the car then the work done by friction is -125,000J

  • 1 month ago

    v² = v₀² + 2ad

    30² = 20² + 2ad

    900 – 400 = 2ad

    ad = 250

    a = 250/d

    F = ma = 500•250/d = 125000/d

    work = Fd = d(125000/d) = 125000 J

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