Question

What change in kinetic energy does an 41000 kg airplane experience on takeoff if it moves a distance of 408.2 m by a constant net force of 4450 newtons?

Answer 1

ΔKE=W

ΔKE=F.d

ΔKE=4450 x 408.2 = 1,82 x 10⁶ J

Answer 2

1 815 236 J

Step-by-step explanation:

Initial Kinetic Energy = 0

For final kinetic energy, we will need the acceleration to find the final velocity

F = ma

4450 = 41 000 a shows a = .10853 m/s^2

xf = vot + 1/2 a t^2

408.2 = 1/2 (.10853)t^2

t = 86.73 seconds

vf = at = .10853 (86.73) = 9.41 m/s

KE final = 1/2 (41000)(9.41 ^2) = 1 815236 J