## Engineering Mechanics: Statics & Dynamics (14th Edition)

$a=0.528m/s^2$
We can determine the required acceleration as follows: $A=\frac{\pi}{4}d^2=\frac{\pi}{4}(0.75)^2=0.4418m^2$ and the flow rate is $Q=v_BA=14\times 0.4418=6.185m^3/s$ The mass flow rate is $m=\rho_a Q=1.22\times 6.185=7.5457Kg/s$ We know that: $\Sigma F_x=m(v_B-v_A)$ $\implies -F=m(v_B-v_A)$ We plug in the known values to obtain: $-F=7.5457-(-14-0)=105.64N$ Now we can calculate the acceleration as $F=ma$ We plug in the known values to obtain: $105.64=200a$ $\implies a=0.528m/s^2$