Question

A block tied between two springs is in equilibrium if upper Spring is cut then the acceleration of the block just after cut is 6m/

Answer 1

How to do it Plz tell

Answer 2

Answer:

Explanation:

let the upper spring be S1 and the lower one be S2..since both the springs are identical so their spring constant will be same = k since the system is kept vertically and the sring constants are same so S2 will be compressed by length x2 and S1 will be elongated by length x1 now balance the force by drawing the FBD kx1 upwards kx2 upwards mg downwards so mg = kx1 + kx2 .....1 therefore mg> kx1 and mg>kx2 so no matter which spring is cut the block will move downwards case I when S1 is cut mg - kx2 = 6m .....2 case II when S2 is cut mg - kx1 = ma ...3 (here a is the acceleration which we are required to find in the question) Adding equation 2 and 3 2mg - (kx1 +kx2) = 6m + ma 2mg - mg = m(6+a) from 1 g=6+a so a = 4 hence the answer is b