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Question: A block of mass \[m\] is connected to two springs of spring constant \[2k\] and \[k\] respectively a...

A block of mass mm is connected to two springs of spring constant 2k2k and kk respectively as shown in vertical plane. At equilibrium both springs are compressed by the same lengths. If suddenly the lower spring is cut what is the acceleration of the block just after spring is cut?

A.2g2g downward
B.2g2g downward
C.2g2g upward
D. none of these

Explanation

Solution

Use the expression for Newton’s second law of motion and formula for spring force. Apply Newton’s second law of motion to the system of block and spring in the vertical direction and determine the value of spring force on upper spring. Apply Newton’s second law of motion to the spring-block system when the lower spring is cut to determine the acceleration of the block.

Formulae used:
The expression for Newton’s second law of motion is
Fnet=ma{F_{net}} = ma …… (1)
Here, Fnet{F_{net}} is net force on the object, mm is mass of the object and aa is acceleration of the object.
The spring force FF is given by
F=kxF = kx …… (2)
Here, kk is spring constant and xx is displacement of the spring.

Complete step by step answer:
We have given that a block of mass mm is attached to two springs of spring constants 2k2k and kk respectively. The free body diagram of the given system of block and springs is as follows:

In the above free body diagram, mgmg is the weight of the block acting in the downward direction, spring force kxkx of upper spring acting in downward direction and spring force 2kx2kx of lower spring acting in upward direction.

At equilibrium, both the springs are compressed by the same length xx.Apply Newton’s second law to the block-spring system in the vertical direction.
kxmg+2kx=0- kx - mg + 2kx = 0
kx=mg\Rightarrow kx = mg
When the lower spring is cut then also the weight of block and spring force of the upper spring will remain the same and this new system will have an acceleration.
Apply Newton’s second law to the new block-spring system in the vertical direction.
kxmg=ma- kx - mg = ma
Substitute mgmg for kxkx in the above equation.
mgmg=ma- mg - mg = ma
a=2g\therefore a = - 2g
Therefore, the acceleration of the block just after the spring is cut is 2g2g in the downward direction.

Hence, the correct option is A.

Note: The students may think how we determined the direction of the acceleration of the block is in downward direction. We can see that the acceleration of the block we determined has a negative sign which shows that the acceleration of the block is in the downward direction.