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Question:

Maximum Deformation of springs?

The four wheels of an automobile of mass 1200 kg are suspended below the body by verticalsprings of constant k7.0×104 N/mIf the forces on all wheels are the same, what will be themaximum deformation of the springs if the automobile is lifted by a crane and dropped on the streetfrom a height of 0.8 m?

Answer:

Assuming all energy of the drop is converted to compression in the springs and none is lost to frictionAssume the mass of the wheels and springs is very small relative to the total mass of the autothe four springs will have a resultant constant of k 28(10^4) N/m PE mgh PS ?kx? where x is the distance of maximum spring compression from the neutral positionin this case we'll let h (x + 0.8) m mgh ?kx? 1200(9.81)(x + 0.8) ?28(10^4)x? 11772x + 9417.6 140000x? 0 140000x? - 11772x - 9417.6 quadratic formula x (11772 ±√(11772? - 4(140000)(-9417.6))) / (2(140000)) x 0.304 m or x -0.221 m as a negative number represents an extension in this reference frame we neglect it x 0.30 m or 30 cm
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