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3 May, 00:43

If this 1300 kg car is driven at 27 m/s through the bottom of a circular dip in the road that has a radius of 600 m, by how much do these springs compress compared to when the car is driven on a flat road

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  1. 3 May, 02:20
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    The springs compress by an extra 12% when driven through the bottom of a circular dip compared to being driven on a flat road.

    Explanation:

    On a flat road, the springs just support the car's weight.

    The weight of the car = force on the springs

    And from Hooke's law, F = kx

    On a flat road,

    k x = mg

    x (flat) = 1300 * 9.8/k = 12740/k

    Through the bottom of the circular dip, the springs not only balance the weight of the car, they also balance the force that keeps it in circular motion.

    kx = (mv²/r) + mg

    kx = (1300 * 27²/600) + (1300*9.8) = 14319.5N

    x (circular dip) = 14319.5/k

    If the value of the springs' spring-constant is available, we'd get straight values for This, but in their absence,

    x (circular dip) / x (flat road) = 14319.5/12740 = 1.12

    The springs compress by an extra 12% when driven through the bottom of a circular dip compared to being driven on a flat road.
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