the ends of a spring are attached to masses 2kg and 3kg. the 3kg block rests on a horizontal surface and 2kg block which is vertically above it is in eqiulibrium producing compression of 1cm of spring initially. find minimum value of distance the 2kg mass must be displaced down so that when it is released the 3kg block may lift off from the ground.

We have two blocks with upper block in equilibrium. The upper block compresses the spring by 1 cm. Therefor force by spring is equal to the weight of 2 kg block. kx = 2g k x 1 = 2g k = (2g)/1 = 20N/cm = 2000N/mNow let the the total compression after exerting force be y. When the force is removed the goes up lifting the lower block upward. At this time the normal reaction force acting on the body becomes zero. This means spring force will be equal to the sum of the weights of upper block and lower block. Ky = 2g + 3gReplacing k by 2000N/m and g by 10m/s2 we get2000y = 2 x 10 + 3 x 102000y = 20 + 302000y = 50y = 50/2000y = 0.025my = 0.025 x 100y = 2.5m