The only force acting on a 2.00 kg body as it moves along a positive x axis has an x component Fx = −kx, where k = 6.00 N/m. The velocity at

Question

The only force acting on a 2.00 kg body as it moves along a positive x axis has an x component Fx = −kx, where k = 6.00 N/m. The velocity at x = 3.00 m is 8.00 m/s. (a) What is the velocity of the body at x = 4.00 m? (b) At what positive value of x will the body have a velocity of 5.00 m/s?

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Acacia 6 months 2021-07-29T02:07:26+00:00 1 Answers 20 views 0

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    2021-07-29T02:08:41+00:00

    Answer:

    a) v at x = 4.0 m is 6.56 m/s

    b) The body will have a velocity of 5.00 m/s at x = 4.69 m

    Explanation:

    Using the work-energy theorem, the work done in moving an object between two points is equal to the change in kinetic energy of the body between those two points.

    W = ΔK.E

    Note that workdone by a position-variable force is given as the definite integral between the two points of F.dx

    W = ∫ F.dx (integral evaluated from point 1 to point 2)

    ΔK.E = (final kinetic energy) – (initial kinetic energy)

    Given

    F = -kx = -6x

    Mass of body = 2.00 kg

    Velocity at x = 3.0 m is 8.0 m/s

    (a) The velocity of the body at x = 4.00 m?

    Workdone by the force moving from point x=3 to x=4

    W = ∫⁴₃ Fdx = ∫⁴₃ (-6x) dx = [-3x²]⁴₃

    = [-3(4²)] – [-3(3²)]

    = -48 – (-27)

    = -21 J

    W = ΔK.E = -21

    ΔK.E = (final kinetic energy) – (initial kinetic energy)

    Final kinetic energy = ?

    Initial kinetic energy at x=3

    = (½)(2)(8²) = 64 J

    ΔK.E = (final kinetic energy) – (initial kinetic energy)

    -21 = (final kinetic energy) – 64

    final kinetic energy = -21 + 64 = 43 J

    Final kinetic energy = (½)(2)(v²) = 43

    v² = 43

    v = 6.557 m/s

    (b) At what positive value of x will the body have a velocity of 5.00 m/s?

    ΔK.E = (final kinetic energy) – (initial kinetic energy)

    Initial kinetic energy at x=3

    = (½)(2)(8²) = 64 J

    Final kinetic energy = (½)(2)(5²) = 25 J

    ΔK.E = 25 – 64 = – 39 J

    W = ΔK.E = – 39 J

    W = ∫ˣ₃ Fdx = ∫ˣ₃ (-6x) dx = [-3x²]ˣ₃

    = [-3(x²)] – [-3(3²)]

    W = -3x² + 27 = – 39

    -3x² = -39-27 = -66

    x² = 22

    x = 4.69 m

    Hope this Helps!!!

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