Physics

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A 75.0kg bobsled is pushed along a horizontal surface by two athletes. After the bobsled is pushed a distance of 4.5m starting from the rest, its speed is 6.0m/s. Find the magnitude of the net force on the bobsled.

  • Physics -

    a= v²/2s
    F=ma

  • Physics -

    Fd=0.5mv^2
    F(4.5)=0.5(75)(6.0)^2
    F=0.5(75)(6.0)^2/4.5
    F=300N

  • Physics -

    Given:
    m=75kg
    d=4.5m
    v=6.00m/s
    angle=180
    Required:
    F=?
    Formrula:
    W=change of KE
    Same as:
    -0.5mv^2=Fdcos()
    Solve for F
    F=-0.5mv^2/dcos()
    F=-0.5(75kg)(6.00m/s)^2/(4.5)(cos 180)
    F=-225/-4.5
    F=50N

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