Question

A block tied between two identical springs is in equilibrium. If upper spring is cut, then the acceleration of the block just after cut is $5m{s}^{-2}$. Now if instead of upper string lower spring is cut, then the acceleration of the block just after the cut will be (Take $g=10m/s^2$),

• A. $1.25m/s^2$
• B. $5m/s^2$
• C. $10m/s^2$
• D. $2.5m/s^2$

Answer 1

The correct option is B $5m/s^2$

Since both the springs are identical, so their spring constant will be same.

Let the spring constant be $k$.

In equilibrium, the upper spring is elongated by $x$ and the lower spring is compressed by $x$. (The elongation and the compression is the same because the springs are in series.)

From the figure,

$mg=kx+kx=2kx$ .....(1)
$\Rightarrow mg>kx$
So the mass will accelerate downwards no matter which spring is cut.

When the upper spring is cut,

$mg-kx=ma$

Given, $a=5m/se{c}^2$

$mg-kx=5m$ ....(2)

Now, when the lower spring is cut

$mg-kx=ma$ ....(3)

Adding equations 2 and 3, we have

$2mg-2kx=5m+ma$

Using equation 1, we have

$2mg-mg=m(5+a)$

$\Rightarrow g=5+a$

$\Rightarrow a=10-5=5m/se{c}^2$

Hence, the correct answer is OPTION B.