A 2.6 kg block is attached to a horizontal rope that exerts a variable force Fx=(20−5x)N, where x is in m. The coefficient of kinetic fricti

Question

A 2.6 kg block is attached to a horizontal rope that exerts a variable force Fx=(20−5x)N, where x is in m. The coefficient of kinetic friction between the block and the floor is 0.25. Initially the block is at rest at x = 0 m.

What is the block’s speed when it has been pulled to x=1.4m?

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Mít Mít 3 years 2021-08-16T01:38:15+00:00 1 Answers 441 views 0

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  1. kimcuc
    0
    2021-08-16T01:39:32+00:00
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    Answer:

    Speed of the block after it is pulled a distance x = 1.4 m is v = 3.28 m/s

    Explanation:

    As per work energy theorem we know that total work done is equal to change in kinetic energy of the system

    so here we have

    \frac{1}{2}mv^2 = \int F.dx - \mu mg x

    so we will have

    \frac{1}{2}(2.6)v^2 = \int (20 - 5x)dx - (0.25)(2.6)(10)(1.4 - 0)

    1.3v^2 = 20(1.4 - 0) - 2.5(1.4^2 - 0) - 0.25(2.6)(10)(1.4)

    v^2 = 10.8

    v = 3.28 m/s

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