Block of 2.5 kg starting from rest falls a vertical distance of 75 cm before striking a vertical colied spring as a result. The spring is compressed at a distance X from its original length when the block comes momentarily to rest the spring constant for the spring is 4000 N/m W what is the distance X that the spring is compressed E sfter the block is momemtarily at reat the spring will return to its equlibruum position and eject the block back into the air what is the speed of the block when it ia at a distance of 50 cm above the end of the spring
Explanation / Answer
we are given with m = 2.5 kg h = 75 cm = 0.75 m k = 4000 N / m as from the conservation of energy weget m g h = (1 / 2) k x2 solve for the distance of compressionx u = 0 m / s s = 50 cm v = ...... m / s v2 - u2 = 2 a s solve for the final velocity solve for the distance of compressionx u = 0 m / s s = 50 cm v = ...... m / s v2 - u2 = 2 a s solve for the final velocity