Suppose the counter attendant pushes a 0.27 kgkg bottle with the same initial speed on a different countertop and it travels 1.7 mm before s

Question

Suppose the counter attendant pushes a 0.27 kgkg bottle with the same initial speed on a different countertop and it travels 1.7 mm before stopping. What is the magnitude of the friction force from this second counter

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Helga 3 years 2021-08-18T15:46:24+00:00 1 Answers 1 views 0

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  1. Euphemia
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    2021-08-18T15:47:33+00:00
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    Answer:

    The force of friction is 622.58 N.

    Explanation:

    Given that,

    Mass of the bottle, m = 0.27 kg

    Finally it stops, v = 0

    Distance traveled by the bottle, d = 1.7 mm = 0.0017 m

    Let the initial speed of the bottle, u = 2.8 m/s

    Let f is the force of friction is acting on the bottle. The force of friction is given by Newton’s second law of motion as :

    -f=ma\\\\a=\dfrac{-f}{m}.............(1)

    Using third equation of motion :

    v^2-u^2=2ad

    v^2=2ad

    v^2=\dfrac{-2fd}{m}\\\\f=\dfrac{-v^2m}{2d}\\\\f=\dfrac{-(2.8)^2\times 0.27}{2\times 0.0017 }

    f = -622.58 N

    So, the force of friction is 622.58 N.

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