Sally asked in Science & MathematicsPhysics · 4 months ago

physics helppp?

If a 65 kg person accelerates uniformly from a standing start to a sprint in 2.7 s over 10.0 m, what was their work done while accelerating?

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  • oubaas
    Lv 7
    4 months ago

    acceleration a = 2d/t^2 = 2*10/2.7^2 = 2.74 m/sec^2

    V^2 = 2*a*d = 2.74*2*10 = 54.9 m^2/sec^2

    Work = KE = 65/2 * 54.9 = 1.8*10^3 joule 

  • 4 months ago

    By s = ut + 1/2at^2

    =>10 = 0 + 1/2 x a x (2.7)^2

    =>a = 2.74 m/s^2

    By W = F x s

    =>W = (ma) x s

    =>W = 65 x 2.74 x 10

    =>W = 1783.26 J

  • 4 months ago

    v² = vo² + 2ad

    v² = 0 + 2a(10)

    v² = 20a   (1)

    v = vo + at

    v = 0 + at

    v = 2,7a

    v² = 2,7²a²   (2)

    (1) in (2)

    20a = 2,7²a²

    7,29a² - 20a = 0

    a1 = 2,74 m/s² (Correct)

    a2 = 0 (It doesn't exist)

    v = 2,7(2,74) = 7,40 m/s

    Ek = 0,5mv²

    Ek = 0,5(65)(7,40²) = 1778,74 j

    Review it!

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