Burning of fuel transfers 4 * 10⁵ W of power into the engine of a 2000-kg vehicle;supposing that the engine is 25% efficient,determine the maximum speed the vehicle can achieve 5 seconds after starting from rest.

Question

Quizplus · Accepted Answer

First, we need to calculate the force exerted by the engine:
Power transferred to the engine = 4F1F1F1S1×F1F1F10 10S1U1P15S1S1P0 W
Efficiency of the engine, η = 25% = 0.25
Power output of the engine, Pout = η × Power transferred to the engine
                                      = 0.25 × 4F1F1F1S1×F1F1F10 10S1U1P15S1S1P0 W
                                      = F1F1F1S1×F1F1F10 10S1U0P375S1S1P0 W

Now, we can use the work-energy principle to find the maximum speed of the vehicle:

Power = Force × Velocity
Pout = Fv
Where v is the velocity of the vehicle

Force, F = ma
Where m is the mass of the vehicle and a is the acceleration of the vehicle

Pout = mav
v = Pout / ma

We can find the acceleration of the vehicle using the work-energy principle:

Work done = Change in kinetic energy
Fds = (1/2)mv^2 - (1/2)m(0)^2
Fds = (1/2)mv^2
a = F/m = (2Fds) / m

Now, substituting the values:

v = Pout / ma
v = F1F1F1S1×F1F1F10 10S1U0P375S1S1P0 / (2Fds)
v = F1F1F1S1×F1F1F10 10S1U0P375S1S1P0 / (2F × d × s)
Taking d = 5 seconds and s = 1000 m
v = (F1F1F1S1×F1F1F10 10S1U0P375S1S1P0 / 2F) × (1/5) × (1/1000)
v = F1F1F1S1×F1F1F1035S1S1P6 m/s
v ≈ 22.4 m/s

Therefore, the maximum speed the vehicle can achieve 5 seconds after starting from rest is approximately 22.4 m/s, which corresponds to choice C.

Burning of Fuel Transfers 4 * 105 W of Power

Burning of Fuel Transfers 4 * 10⁵ W of Power