Thrust force due to ejection of water on the cart is counter balanced by spring force on the cart
so we can write
[tex]F_t = v \frac{dm}{dt}[/tex]
here we have
[tex]\frac{dm}{dt}[/tex] = rate of mass ejection
[tex]\frac{dm}{dt} = \rho a\frac{dx}{dt} = \rho a v[/tex]
now from above formula
[tex]F_t = \rho a v(v) = \rho av^2[/tex]
Now by force balance equation
[tex]F_t = F_s[/tex]
[tex]\rho a v^2 = kx[/tex]
by rearranging the terms now
[tex]v= \sqrt{\frac{kx}{\rho a}}[/tex]
