Question

the ends of springs are attached to a block of mass 3 kg and 2kg 3 kg block rests on a horizontal surface and 2 kg block which is vertically above it is in equilibrium process producing a compression of 1 cm of the string the value of length to which the 2 kg must be compressed so that when it is released 3 kg block may be lifted Off The Ground is​

Answer 1

Answer:

0.025m.

Explanation:

Since, we know that the two system are in equilibrium, hence when the spring is compressed by 1 cm, the force(for 2kg block) will be given as kx which will be 2g=k*1 which on solving we will get k=20*100 = 2000 N/m.

So, when the blocks are in equilibrium at the time of releasing the 3kg block then the compression will be x1. So, kx1 will be 2g+3g which we will get on substituting the value of k. 2000x1=20+30 which on solving we will get the value of x1 = 0.025 m.