Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving with a speed

Question

Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m? b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

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Thiên Hương 6 months 2021-07-29T23:51:22+00:00 1 Answers 57 views 0

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    2021-07-29T23:53:12+00:00

    Correct question:

    Consider the motion of a 4.00-kg particle that moves with potential energy given by

     U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}

    a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

    b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

    Answer:

    a) 3.33 m/s

    b) 0.016 N

    Explanation:

    a) given:

    V = 3.00 m/s

    x1 = 1.00 m

    x = 5.00

     u(x) = \frac{-2}{x} + \frac{4}{x^2}

    At x = 1.00 m

     u(1) = \frac{-2}{1} + \frac{4}{1^2}

    = 4J

    Kinetic energy = (1/2)mv²

     = \frac{1}{2} * 4(3)^2

    = 18J

    Total energy will be =

    4J + 18J = 22J

    At x = 5

     u(5) = \frac{-2}{5} + \frac{4}{5^2}

     = \frac{4-10}{25} = \frac{-6}{25} J

    = -0.24J

    Kinetic energy =

     \frac{1}{2} * 4Vf^2

    = 2Vf²

    Total energy =

    2Vf² – 0.024

    Using conservation of energy,

    Initial total energy = final total energy

    22 = 2Vf² – 0.24

    Vf² = (22+0.24) / 2

     Vf = \sqrt{frac{22.4}{2}

    = 3.33 m/s

    b) magnitude of force when x = 5.0m

     u(x) = \frac{-2}{x} + \frac{4}{x^2}

     \frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}

     = \frac{2}{x^2} - \frac{8}{x^3}

    At x = 5.0 m

     \frac{2}{5^2} - \frac{8}{5^3}

     F = \frac{2}{25} - \frac{8}{125}

    = 0.016N

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