A 20000 kg airplane takes off by traveling 300 m down a runway while the engines exert a continuous force of 4000 N.(a) Calculate the change in kinetic energy of the airplane during takeoff.(b) Determine the speed of the airplane as it leaves the runway.
Work in = change in Ke = force * distance
= 4000 * 300 = 12*10^5
so
(1/2)(2*10^5) v^2 = 12*10^5
v^2 = 12
v = 2 sqrt3
4,000 N is not enough to get this plane off the ground.
To calculate the change in kinetic energy (ΔKE) of the airplane during takeoff, we need to use the formula:
ΔKE = KE_final - KE_initial
To find the initial kinetic energy (KE_initial), we need the initial speed (v_initial) of the airplane, which is assumed to be zero since it starts from rest.
KE_initial = (1/2) * m * v_initial^2
Given that the mass (m) of the airplane is 20000 kg and the initial speed (v_initial) is zero, we can calculate KE_initial:
KE_initial = (1/2) * 20000 kg * (0 m/s)^2 = 0 J
To find the final kinetic energy (KE_final), we need the final speed (v_final) of the airplane. We can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy:
Work = ΔKE
The work (W) done on the airplane is equal to the product of the force applied (F) and the displacement (s) traveled by the airplane:
W = F * s
Given that the force (F) applied by the engines is 4000 N and the displacement (s) is 300 m, we can calculate the work done on the airplane:
W = 4000 N * 300 m = 1,200,000 J
Using the work-energy principle, we can set the work equal to the change in kinetic energy to find KE_final:
W = ΔKE
1,200,000 J = ΔKE
Therefore, the change in kinetic energy of the airplane during takeoff is 1,200,000 J.
To determine the speed of the airplane as it leaves the runway (v_final), we can use the equation for final kinetic energy:
KE_final = (1/2) * m * v_final^2
Plugging in the given mass (m) of 20000 kg and the calculated change in kinetic energy (ΔKE) of 1,200,000 J, we can solve for v_final:
1,200,000 J = (1/2) * 20000 kg * v_final^2
Rearranging the equation and solving for v_final, we get:
v_final = √(2 * ΔKE / m)
v_final = √(2 * 1,200,000 J / 20000 kg) = √(120 J/kg) ≈ 10.95 m/s
Therefore, the speed of the airplane as it leaves the runway is approximately 10.95 m/s.