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Work Done By A Force Acting on The Piston

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:

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.
$$ W=\frac{1}{2}\frac{\rho V^3}{s^2t^2}$$

October 15, 2017

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