You are asked to design spring bumpers for the walls of a parking garage. A freely rolling 1300 kg car moving at 0.62 m/s is to compress the

Question

You are asked to design spring bumpers for the walls of a parking garage. A freely rolling 1300 kg car moving at 0.62 m/s is to compress the spring no more than 0.072 m before stopping. What should be the force constant of the spring? Assume that the spring has negligible mass.

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Thu Giang 4 years 2021-08-27T19:15:22+00:00 1 Answers 78 views 0

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  1. Calantha
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    2021-08-27T19:17:17+00:00
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    Answer:

    F=3470.2778\ N

    Explanation:

    Given:

    mass of the car, m=1300\ kg

    parking speed of the car, u=0.62\ m.s^{-1}

    compression of spring bumpers on the walls, \delta x=0.072\ m

    Using the equation of motion:

    v^2=u^2+2.a.\delta x

    where:

    v= final speed of the car =0\ m.s^{-1}

    a= acceleration of the car while compressing the spring (will be -ve since final velocity tends to zero)

    0^2=0.62^2+2\times a\times 0.072

    a=-2.6694\ m.s^{-1} (negative sign denotes that it is reducing the speed )

    Now the force:

    F=m.a

    F=1300\times 2.6694

    F=3470.2778\ N

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