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Yahoo!

  • 3 years ago

P.E of a particle free to move to along x-axis is given by (x^3 - x^2/2)J where x is in metre . The particle is initially kept at x=0 and then given a sight displacement x in +ve direction. The force acting on the particle will displace it to a Distance ?

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  1. Yahoo!
    • 3 years ago
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    @experimentX @JFraser @siddhantsharan @gerryliyana

  2. Yahoo!
    • 3 years ago
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    @ghazi

  3. experimentX
    • 3 years ago
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    F = - d(PE)/dx

  4. Yahoo!
    • 3 years ago
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    - (3x^2 - x)

  5. experimentX
    • 3 years ago
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    F = 0 ... find the equilibrium

  6. Yahoo!
    • 3 years ago
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    x = 1/3

  7. Yahoo!
    • 3 years ago
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    Thxx.)

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