A force on a particle depends on position such that F(x) = (3.00 N/m2)x2 + (6.00 N/m)x for a particle constrained to move along the x-axis.

Question

A force on a particle depends on position such that F(x) = (3.00 N/m2)x2 + (6.00 N/m)x for a particle constrained to move along the x-axis. What work is done by this force on a particle that moves from x = 0.00 m to x = 2.00 m?

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Thiên Ân 2 weeks 2021-08-29T02:58:31+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-29T02:59:35+00:00

    Answer:

    The work done by a particle from x = 0 to x = 2 m is 20 J.

    Explanation:

    A force on a particle depends on position constrained to move along the x-axis, is given by,

    F(x)=(3\ N/m^2)x^2+(6\ N/m)x

    We need to find the work done on a particle that moves from x = 0.00 m to x = 2.00 m.

    We know that the work done by a particle is given by the formula as follows :

    W=\int\limits {F{\cdot} dx}

    W=\int\limits^2_0 {(3x^2+6x){\cdot} dx} \\\\W={(x^3}+3x^2)_0^2\\\\\W={(2^3}+3(2)^2)\\\\W=20\ J

    So, the work done by a particle from x = 0 to x = 2 m is 20 J. Hence, this is the required solution.

    0
    2021-08-29T02:59:59+00:00

    Answer:

    Explanation:

    Force is given by

    F = 3x² + 6x

    Particle moves from x = 0 m to 2 m

    Work done is

    W=\int_{x_{1}}^{x_{2}}F(x)dx

    W=\int_{0}^{2}\left ( 3x^{2}+6x \right )dx

    W=\left ( x^{3}+3x^{2} \right )_{0}^{2}

    W = 8 + 12 – 0 – 0

    W = 20 J

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