The energy of compressed crossbow is `U = (kx^2)/2` , where k = 6000 N/m and x is the displacement. When crossbow goes from compressed state to relaxed state, its does work equal to the change of its potential energy from U to 0. So, the work done by crossbow as it goes from a state with 0.2 m displacement to relaxed state is `W = (kx^2)/2` , where x = 0.2 m.
Power is the rate at which the work is done:
`P = (DeltaW)/(Deltat)` , where `Deltat ` is the time interval and `DeltaW` is the work done during this time interval. The average power output of the crossbow is then
`P = ((kx^2)/2)/(Deltat)` , for `Delta t = 0.1 s`
`P = ((6000 * 0.2^2)/2)/0.1 = 1,200 W` (Watts.)
The average power output of the crossbow is 1,200 Watts.