Question

An aeroplane of mass 5000 kg is flying at a height 1 kilometre.At a position‚ the potential energy and kinetic energy are equal. Calculate kinetic energy and velocity.​

Answer 1

Answer:

The velocity is 140 m/s.

Step-by-step explanation:

Mass of aeroplane = 5000 kg

Flying at height of = 1 km

At a point above the ground where,

Potential Energy = Kinetic Energy

Gravitational Potential Energy :

$\longrightarrow \: \sf{PE_g = mgh}$

• m = 5000 mg
• g = 9.8 m/s²
• h = 1 km = 1000 m

$\longrightarrow \: \sf{PE_g =5000 \times 9.8 \times 1000}$

$\longrightarrow \: \sf{PE_g =5000 \times 9800}$

$\longrightarrow \: \sf{PE_g = 49000000}$

Gravitational Potential Energy = 49000000 J

Kinetic Energy = 49000000 J

$\rule{300}{1.5}$

Velocity:

$\sf{\longrightarrow} \: v = \sqrt{ \dfrac{KE}{ \frac{1}{2}m}}$

• KE = 49000000 J
• m = 5000 kg

$\sf{\longrightarrow} \: v = \sqrt{ \dfrac{49000000}{ \frac{1}{2} \times 5000}}$

$\sf{\longrightarrow} \: v = \sqrt{ \dfrac{49000000}{2500}}$

$\sf{\longrightarrow} \: v =\dfrac{7000}{50}$

$\sf{\longrightarrow} \: v = 140$

Velocity = 140 m/s

Therefore, the velocity is 140 m/s.

Answer 2

Mass of aeroplane = 5000 kg

Height = 1000 m (h)

KE + PE = constant

⇒ 2PE(f) = PE(i)

⇒ KE = ½ × mgh

⇒ ½ mv^2 = ½ mgh

⇒ v^2 = gh

⇒ v = √gh = √10 m/s^2 × 1000 m

⇒ v = 100 m/s

So, the point where PE = KE's velocity is 100 m/s.