A single force acts on a 4.0 kg particle-like object in such a way that the position of the object as a function of time is given by x = 3.0

Question

A single force acts on a 4.0 kg particle-like object in such a way that the position of the object as a function of time is given by x = 3.0t − 4.0t 2 + 1.0t 3, with x in meters and t in seconds. Find the work done on the object by the force from t = 0 to t = 8.0 s.

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bonexptip 1 month 2021-08-06T05:48:59+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-06T05:50:11+00:00

    Answer:

    31.232 kJ

    Explanation:

    Given,

    mass of the particle, m = 4 Kg

    Position of object as the function of time

    x = 3.0 t - 4.0 t^2 + 1.0 t^3

    Work done by the object by the force from t = 0 to t = 8.0 s =?

    we know,

    v = \dfrac{dx}{dt}

    v =3-8t +3 t^2

    velocity at t = 0 s

    u = 3 – 8 x 0 + 3 x 0 = 3 m/s

    velocity at t= 8 s

    v = 3 – 8 x 8 + 3 x 8 x 8

    v = 125 m/s

    Work done is equal to change in KE

    W = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2  

    W = \dfrac{1}{2}m(v^2 -u^2)

    W = \dfrac{1}{2}\times 4\times (125^2 -3^2)

    W = 31.232 k J

    Hence, work done is equal to 31.232 kJ

    0
    2021-08-06T05:50:47+00:00

    Answer:

    34304 Joule

    Explanation:

    mass of particle, m = 4 kg

    x = 3t – 4t² + t³

    Let v is the velocity

    v = dx/dt = 3 – 8t + 3t²

    Let a is the acceleration

    a = dv/dt = – 8 + 6t

    Work is defined as the product of force.

    \int dW=\int madx

    W=4\times \int _{0}^{8} \left ( -24t+82 t - 72t^{2}+18t^{3} \right )dx

    W=4\times \left ( -24t+41t^{2} - 24t^{3}+4.5t^{4}\right )_{0}^{8}

    W=4\times \left ( -24\times 8+41\times 64 - 24\times 512+4.5\times 4096\right )

    W = 4 x (- 192 + 2624 – 12288 + 18432)

    W = 34304 Joule  

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