A small plane mass 600 kg lifts off from the runway at a speed of 145 Km/Kr. 2 Minutes later, the plane has reached an altitude of 300 meters and its airspeed has dropped to 90 Km/Hr. Assume the power output of the engine was constant during this time. Ignore air resistance and compute the HP (horsepower) produced by the engine.
Divide total energy gain by 120 seconds for the energy of power delivered by the engine.
P.E. gain = M g H = 1.746*10^6 J
K.E. loss = (M/2)(V1^2 - V2^2)
V1 = 40.28 m/s
V2 = 25.00 m/s
K.E. loss = 2.992*10^5 J
Net energy gain = 1.465*10^6 J
Power during interval = 1.465*10^6/120 = 12210 W = 164 horsepower
This is an underestimate because there is also air friction drag that the engine must overcome, and because the propellers are not 100% efficient
To compute the horsepower (HP) produced by the engine, we need to calculate the amount of work done in lifting the plane to its current altitude in the given time interval.
First, let's convert the plane's airspeed from km/hr to m/s. We know that:
1 km/hr = 1000 m / 3600 s = 5/18 m/s
So, the initial airspeed of 145 km/hr would be:
145 km/hr * 5/18 m/s = 40.28 m/s
And the final airspeed of 90 km/hr would be:
90 km/hr * 5/18 m/s = 25 m/s
Next, we need to calculate the change in kinetic energy and potential energy of the plane.
The change in kinetic energy is given by the formula:
ΔKE = (1/2) * m * (vf^2 - vi^2)
Where:
ΔKE = change in kinetic energy
m = mass of the plane
vf = final velocity
vi = initial velocity
Substituting the values, we get:
ΔKE = (1/2) * 600 kg * (25 m/s)^2 - (40.28 m/s)^2
Simplifying this, we find:
ΔKE = -342800 Joules
Since there is no air resistance, all the work done is converted into potential energy. The change in potential energy is given by the formula:
ΔPE = m * g * Δh
Where:
ΔPE = change in potential energy
m = mass of the plane
g = acceleration due to gravity, approximately 9.8 m/s^2
Δh = change in altitude
Substituting the values, we get:
ΔPE = 600 kg * 9.8 m/s^2 * 300 m
Simplifying this, we find:
ΔPE = 1764000 Joules
The work done by the engine is equal to the sum of the change in kinetic energy and the change in potential energy. Therefore, the total work done is:
Work = ΔKE + ΔPE
= -342800 J + 1764000 J
= 1421200 J
Finally, we can calculate the power output of the engine using the formula:
Power = Work / Time
Since the time interval is given as 2 minutes, we need to convert it to seconds:
2 minutes * 60 seconds/minute = 120 seconds
Substituting the values, we find:
Power = 1421200 J / 120 s
= 11843.33 watts
To convert watts to horsepower, we know that:
1 horsepower (HP) = 745.7 watts
Therefore, the power output of the engine is:
Power = 11843.33 watts / 745.7
≈ 15.89 HP
So, the engine produced approximately 15.89 horsepower.