A block of mass 0.251 kg is placed on top of a light vertical spring of force co

Question

A block of mass 0.251 kg is placed on top of a light vertical spring of force constant 4850 N/m and pushed downward so that the spring is compressed 0.092 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise? Enter your answer to 2 decimal places. _______m

Explanation / Answer

By PE(spring) = PE(gained by block) =>1/2kx^2 = mgh =>h = kx^2/2mg =>h = [4850 x (0.092)^2]/[2 x 0.251 x 9.8] =>h = 8.28m Thus the net height above the point of release(H) = h - x =>H = 8.28 - 0.092 = 8.188m

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