Question

When its 100-hp engine is generating full power, a small airplane with mass 700 kggains altitude at a rate of 2.5 m/s. What fraction of the engine power is being used to make theairplane climb? (The remainder is used to overcome the effects of air resistance and of inefficienciesin the propeller and engine. Assumeg= 9.8 m/s2.)

Answer 1

For the given problem it is necessary to recap the concepts about Power, that is, is the rate of doing work or of transferring heat, i.e. the amount of energy transferred or converted per unit time.

The equation for power can be written as

$P = FV$

Where,

F= Net Force

V =Velocity

By the second newton law, force can be:

F = mg

Where m means the mass and g the gravity acceleration.

We can also write the equation as,

P = mgv

Replacing the values

P = 700*9.8*25

P = 17.15kW

Tenemos unidades en dos sistemas diferentes, por lo que convertimos los HP a Sistema internacional y tenemos que

1hp = 0.746kW

Then the fraction $\eta$is,

$\eta = \frac{17.15kW}{0.746kW}$

$\eta = 22.989$

Therefore the fraction of the engine power is being used to make the airplane climb is 22.984%