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The potential energy of a particle is de...

The potential energy of a particle is determined by the expression `U=alpha(x^(2)+y^(2))`, where `alpha` is a positive constant. The particle begins to move from a point with the co-ordinates (3, 3) only under the action of potential fields force. When it reaches the point (1, 1) its kinetic energy is `4 Kalpha`. Find the value of K.

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Knowledge Check

  • The potential energy of a particle is determined by the expression U=alpha(x^2+y^2) , where alpha is a positive constant. The particle begins to move from a point with coordinates (3, 3) , only under the action of potential field force. Then its kinetic energy T at the instant when the particle is at a point with the coordinates (1,1) is

    A
    (a) `8alpha`
    B
    (b) `24alpha`
    C
    (c) `16alpha`
    D
    (d) Zero

  • The potential energy of a particle with displacement X is U(X) . The motion is simple harmonic, when (K is a positive constant)

    A
    `U=-(KX^2)/(2)`
    B
    `U=KX^2`
    C
    `U=K`
    D
    `U=KX`

  • The potential energy of a particle in a field is U= (a)/(r^2)-b/r , where a and b are constant. The value of r in terms of a and b where force on the particle is zero will be:

    A
    `a/b`
    B
    `b/a`
    C
    `(2a)/(b)`
    D
    `(2b)/(a)`

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