a particle of mass m=375 g is launched with velocity of A =4 moves with a length AB=2.5m under the action of tractive force F=3.46 N making

Question

a particle of mass m=375 g is launched with velocity of A =4 moves with a length AB=2.5m under the action of tractive force F=3.46 N making an angle 30. during its motion between A and B which is subjected to a frictional force f=1.5 N
calculate V of B by applying the kinetic energy theorem​

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Verity 4 years 2021-07-19T14:54:18+00:00 1 Answers 17 views 0

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  1. binhan
    0
    2021-07-19T14:55:26+00:00
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    Answer:

    The final speed is 5.78 m/s.

    Explanation:

    mass, m = 375 g = 0.375 kg

    initial velocity, u = 4 m/s

    Distance, s = 2.5 m

    Angle, A = 30 degree

    Force, F = 1.5 N

    let the final velocity is v.

    Use the work energy theorem

    Work done = change in kinetic energy

    W= 0.5 m(v^2 - u^2)\\\\F s cos A= 0.5 m (v^2 - u^2)\\\\1.5\times 2.5\times cos30= 0.5\times 0.375\times (v^2 - 16)\\\\v = 5.78 m/s

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