A block of mass 2.5kg, starting from rest, falls a verticaldistance of 75cm befo

Question

A block of mass 2.5kg, starting from rest, falls a verticaldistance of 75cm before striking a vertical coiled spring. As aresult of this contact the spring is compressed a distance, x, fromits original length when the block comes mometarily to rest. Thespring constant for the spring is 4000N/m. What is the distance x, that the spring is compressed? After the block is momentarily at rest, the spring will returnto its equilibrium position and eject the block back into the air.What is the speed of the block when it is a distance of 50cm abovethe end of the spring? A block of mass 2.5kg, starting from rest, falls a verticaldistance of 75cm before striking a vertical coiled spring. As aresult of this contact the spring is compressed a distance, x, fromits original length when the block comes mometarily to rest. Thespring constant for the spring is 4000N/m. What is the distance x, that the spring is compressed? After the block is momentarily at rest, the spring will returnto its equilibrium position and eject the block back into the air.What is the speed of the block when it is a distance of 50cm abovethe end of the spring?

Explanation / Answer

Given that the mass of block is m = 2.5 kg        Initial heightabove spring is h = 75 cm = 0.75 m        Spring constant is k =4000 N/m ---------------------------------------------------------------------------    Apply conservation of energy at the height h andat compressed length X from equilibrium position                            mgh = (1/2)kX2 - mgX X2 - 2gX / k -2gh /k =0 X2 - (0.0049 ) X -0.00367 = 0    Then we get X = 0.063 m    Apply conservation of energy at compressedlength X and at equilibrium position                            (1/2)kX2 = (1/2)mV2 +mgx
                                     V = ( kX2 / m - 2gx )
                                           =-------- m/s Then the velocity at the height h =0.5 m is                 (1/2)mV2 = mgh +(1/2)mu2                             V2 = 2gh + u2              slovefor u = ---------- m/s

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