Show that a 2,500,000-J change in kinetic energy occurs for an airplane that is moved 500 m in takeoff by a sustained force of 5000 N.

The answer to your question is given below

Explanation:

To solve this problem, we'll assume that the plane is initially at rest.

Hence, the kinetic energy of the plane at rest is zero i.e Initial kinetic energy (KE₁) = 0

Next, we shall determine the final kinetic energy of the plan when the force was applied. This can be obtained as follow:

Force (F) = 5000 N

Distance (s) = 500 m

Energy (E) =?

E = F × s

E = 5000 × 500

E = 2500000 J

Since energy an kinetic energy has the same unit of measurement, thus, the final kinetic energy (KE₂) of the plane is 2500000 J

Finally, we shall determine the change in the kinetic energy of the plane. This can be obtained as follow:

Initial kinetic energy (KE₁) = 0

Final kinetic energy (KE₂) = 2500000 J

Change in kinetic energy (ΔKE) =?

ΔKE = KE₂ – KE₁

ΔKE = 2500000 – 0

ΔKE = 2500000 J

Hence, the change in the kinetic energy of the plane is 2500000 J.

double the distance = divide strength by 4

the ima (higher than 1m) will be greater than the ima with the height of 1m.