# How do I find the speed of a block supported between two springs?

#### Chemist116

The problem is as follows:

The figure from below shows a block whose mass is $10\,kg$. The block is held by two springs on each side. Each spring has the same deformation constant. Initially each spring are not deformed. Find the deformation in cm of the spring such the maximum speed of the block is $2\,\frac{cm}{s}$. Consider the friction on the surface is negligible.

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&4\,cm\\ 2.&2\,cm\\ 3.&3\,cm\\ 4.&1\,cm\\ 5.&5\,cm\\ \end{array}$

What I attempted to do to solve this problem was to use the conservation of mechanical energy as follows:

In the left side of the spring it would be:

$\frac{1}{2}kx^2+\frac{1}{2}mv^2=-\frac{1}{2}kx^2$

For this part I assumed that when the spring is compressed its energy is positive which I think is happening in the left side so it will be storing energy while in the right side the other spring will be releasing energy hence negative?. Can somebody help me with these assumptions?

$\frac{1}{2}mv^2=-\frac{1}{2}kx^2-\frac{1}{2}kx^2$

$\frac{1}{2}mv^2=-kx^2$

At this point I realized that I can't take the square root of a negative number so I guessed that it might be positive, but I don't know the justification.

$x=\sqrt{\frac{mv^2}{2k}}=\sqrt{\frac{10\left(2\right)^2}{2 \times 20}}=\sqrt{1}=1\,cm$

This correspond to the fourth alternative and it checks with the answers sheet but I'm not very convinced of my method. Can somebody clear my doubts regarding this problem?.

Last edited:

#### skeeter

Math Team
stretch or compression, the sign of elastic potential energy is > 0

one spring displaced $x \implies$ the other spring is displaced $-x$ ...

total initial spring energy at max displacement from equilibrium = max kinetic energy at equilibrium (no deformation of the spring)

$\dfrac{1}{2}k(-x)^2 + \dfrac{1}{2}k(x)^2 = \dfrac{1}{2}mv^2$

$2kx^2 = mv^2 \implies x = \pm\sqrt{\dfrac{mv^2}{2k}} = \pm \sqrt{\dfrac{10 \text{ kg} \cdot (0.02 \text{ m/s})^2}{2 \cdot 10 \text{ kg/sec}^2}} \approx \pm 0.01 \text{ m} = \pm 1 \text{ cm}$

... which basically says one may displace the mass to the right or left of equilibrium to achieve the same maximum speed.