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Let’s discuss a problem and find out the work done by a force acting on the piston. First, try the problem and then read the solution.

The Problem:

A cylinder filled with water of volume (V) is fitted with a piston and placed horizontally. There is a hole of cross-sectional area (s) at the other end of the cylinder, (s) being much smaller than the cross-sectional area of the piston. Show that the work to be done by a constant force acting on the piston to squeeze all water from the cylinder in time (t) is given by $$W=\frac{\rho V^3}{2s^2t^2}$$ where (\rho) is the density of water.

Solution:

The volume of water flowing out per second $$Q=sv$$
where (v) is the speed of sound and (s) is the cross-sectional area.
Volume flowing out $$V=Qt=svt$$ $$\frac{ \rho v^2}{2}=P$$$$= \frac{F}{A}$$ $$=\frac{FL}{AL}$$ $$=\frac{W}{V}$$
where (L) is the length of the cylinder and (W) is the work done.
$$W=\frac{1}{2}\frac{\rho V^3}{s^2t^2}$$