energy conservation?

Question

An 865 kg airplane starts at rest on an airport runway at sea level.
(a) What is the change in mechanical energy of the airplane if it climbs to a cruising altitude of 2400 m and maintains a constant speed of 95.0 m/s?
(b) What cruising speed would the plane need at this altitude if its increase in kinetic energy is equal to its increase in potential energy?

Answer 1

(a) Mechanical Energy = Kinetic energy + Potential energy
............... ..................= ½ m v² + mgh
Mass of the air craft = 865 Kg
Velocity of the aircraft = 95 m/s
Height of the aircraft = 2400 m
Total Mech. Energy = ½ m v² + m g h
.............. ...................= ½ x 865 x 95² + 865 x 9.8 x 2400
............... ..................= 3903312.5 + 20344800
.................. ................= 24248112.5 J
................... ..............= 24248.1125 KJ or 24.2481125 MJ
................................. ..============= =============
(b) Increase in PE = 20344800 J
To have this much of KE the aircraft must have a velocity "v"
½ m v² = 20344800 = ½ x 865 x v²
,........v² = (20344800 x 2) / 865 = 47040
......... v = √47040 = 216.89 m/s
........................ .....===========
If the aircraft attains the velociy "216.89 m/s" her KE will be the same as its PE.

Answer 2

There is a change in poitential energy from 0 at sea level to:

PE = mgh where h = cruising altitude

It has kinetic energy

KE = 1/2mv^2 where v = cruising speed

Total energy is E =PE +KE

For part b set

1/2mv2^2 = mgh and solve for v2

v2 = sqrt(2gh)

I'll let you plug in the numbers

Answer 3

change in ME = (KE_final + PE_final) - (KE_init + PE_init)
KE_final = .5 * 865 * (95m/s)^2
PE_final = mgh = 865 * 9.8 * 2400
KE_initial = 0
PE_initial = 0

Part B
--------------------
mgh = .5 * m * v^2 | v = cruising speed