## College Physics (4th Edition)

The average force on the jet due to the catapult is $6.97\times 10^5~N$
We can convert the final speed to units of $m/s$: $v_f = 160~mi/h\times \frac{1609~m}{1~mi} \times \frac{1~h}{3600~s} = 71.5~m/s$ We can find the plane's acceleration: $v_f^2 = v_0^2+2ax$ $a = \frac{v_f^2 - v_0^2}{2x}$ $a = \frac{(71.5~m/s)^2 - 0}{(2)(90~m)}$ $a = 28.4~m/s^2$ We can convert the force of each engine to units of newtons: $27,000~lb\times \frac{4.448~N}{1~lb} = 120,096~N$ We can find the force $F_c$ of the catapult: $\sum F = ma$ $F_c+(2)(120,096~N) = (33,000~kg)(28.4~m/s)$ $F_c = (33,000~kg)(28.4~m/s)- (2)(120,096~N)$ $F_c = 6.97\times 10^5~N$ The average force on the jet due to the catapult is $6.97\times 10^5~N$