Question

A pistol shrimp squirts a bullet of water from it's claw to stun its prey. The water bullet travels at 25m/s. In order to do this it springs back its claw like a pistol Assuming the water bullet has a mass of 10 g and the shrimp's muscles stretches a total of 2 cm, what would be the spring constant of the shrimp’s muscle? Round answer to the nearest whole number and include the proper units.

Answer 1

The spring constant of shrimp's muscle is 15625 N/m.

Given data:

The velocity of the water bullet is v=25 m/s.

The mass of water bullet is m=10 g.

The stretche in the muscle is x=2 cm.

The amount of elastic potential energy stored in the muscle will be equal to the kinetic energy of the water bullet. It can be written as,

$\begin{gathered} PE=KE \\ \frac{1}{2}kx^2=\frac{1}{2}mv^2 \\ k=\frac{mv^2}{x^2} \\ k=\frac{(10g\times\frac{1kg}{1000\text{ g}})(\frac{25m}{s})^2}{(2cm\times\frac{1\text{ m}}{100cm})^2} \\ k=\frac{15625kg}{s^2}^{}\times\frac{1\text{ N/m}}{\frac{1kg}{s^2}} \\ k=15625\text{ N/m} \end{gathered}$

Thus, the spring constant of shrimp's muscle is 15625 N/m.