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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.

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christian

  1. a= v²/2s
    F=ma

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    Elena

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

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    Anonymous

  4. 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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    Anonymous

  5. Wnet = 1/2mv^2 (Final) - 1/2mv^2 (Initial)
    Initial = 0 because the sled started from rest, so we first need to find the final.
    KE = 1/2mv^2
    m = 75kg
    v = 6m/s
    KE = 1/2 (75)(36)^2 = 1350j
    1350j is our energy, to find force we use
    E = Fd so F= E/d
    d = 4.5m
    F= 1350/4.5 = 300N
    300N is our final force, and we already know our initial force is 0
    So, using Wnet = 1/2mv^2 (Final) - 1/2mv^2 (Initial) we can figure out our net force
    Wnet = 300N - 0N
    Wnet = 300N

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    Quinn

  6. Wnet = 1/2mv^2 (Final) - 1/2mv^2 (Initial)
    Initial = 0 because the sled started from rest, so we first need to find the final.
    KE = 1/2mv^2
    m = 75kg
    v = 6m/s
    KE = 1/2 (75)(36)^2 = 1350j
    1350j is our energy, to find force we use
    E = Fd so F= E/d
    d = 4.5m
    F= 1350/4.5 = 300N
    300N is our final force, and we already know our initial force is 0
    So, using Wnet = 1/2mv^2 (Final) - 1/2mv^2 (Initial) we can figure out our net force
    Wnet = 300N - 0N
    Wnet = 300N

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    ZZZoZZZ

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