A 1.50 multiplied by 103 kg car starts from rest and accelerates uniformly to 19.9 m/s in 11.6 s. Assume that air resistance remains constant at 400 N during this time.
(a) Find the average power developed by the engine.
1 W
(b) Find the instantaneous power output of the engine at t = 11.6 s, just before the car stops accelerating
(a) Paverage*t = Kinetic energy increase + work done against air friction.
KE increase = 297,000 J
t = 11.6 s
distance traveled = X = (19.9/2)*11.6 = 115.4 m
The work done against friction = 400N*115.4m = 46,170 J
Pav*t = 343,170 J
Pav = 29,584 W
The assumption of constant air friction is completly unrealistic and will underestimate the final instantaneous power level.
You can come up with an answer by using the instantaneous relation
P = (Force)*(Velocity)
For the force, add the air resistance to the force required to accelerate,
F @ t=11.6s = M*a + 400 N
= 2573 + 400 = 2973 N
Instantaneous power = F*V = 59,168 W
To find the average power developed by the engine, we need to find the work done by the engine and divide it by the time taken. The work done by the engine is equal to the change in kinetic energy of the car.
Given:
Mass of the car (m) = 1.50 x 10^3 kg
Initial velocity (u) = 0 m/s
Final velocity (v) = 19.9 m/s
Time taken (t) = 11.6 s
Air resistance (R) = 400 N
(a) Average power developed by the engine:
The change in kinetic energy of the car can be calculated using the work-energy theorem:
ΔKE = KE_final - KE_initial
= (1/2)mv^2 - (1/2)mu^2
Substituting the given values:
ΔKE = (1/2)(1.50 x 10^3 kg)(19.9 m/s)^2 - (1/2)(1.50 x 10^3 kg)(0 m/s)^2
= 298,350 J
The average power developed by the engine can be calculated using the formula:
Average power = Work done / Time taken
P_avg = ΔKE / t
= 298,350 J / 11.6 s
≈ 25,723.28 W
≈ 25.7 kW (rounded to one decimal place)
Therefore, the average power developed by the engine is approximately 25.7 kW.
(b) Instantaneous power output of the engine at t = 11.6 s:
The instantaneous power output at any time during the acceleration can be calculated by multiplying the force applied by the engine with the instantaneous velocity of the car.
The net force acting on the car during acceleration is the sum of the force applied by the engine and the air resistance:
Net force (F_net) = Force_ engine - Air resistance
At t = 11.6 s, just before the car stops accelerating, the net force is only due to air resistance as the engine force becomes zero. Therefore:
F_net = - Air resistance
= - 400 N
The instantaneous power output can be calculated using the formula:
Instantaneous power = Force x Velocity
P_instantaneous = F_net x v
= - 400 N × 19.9 m/s
= - 7,960 W
≈ - 7.96 kW (rounded to two decimal places)
Therefore, the instantaneous power output of the engine at t = 11.6 s, just before the car stops accelerating, is approximately -7.96 kW.
To find the average power developed by the engine, we know that power is defined as the rate of doing work. The equation for power is:
Power = Work / Time
In this case, we need to find the work done by the engine over the given time interval. The work done by the engine can be calculated using the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy. The equation for work done is:
Work = (1/2) * mass * (final velocity^2 - initial velocity^2)
In this case, the initial velocity is 0 m/s (since the car starts from rest), the final velocity is given as 19.9 m/s, and the mass of the car is given as 1.50 multiplied by 103 kg. Plugging in these values into the equation, we get:
Work = (1/2) * (1.50 * 10^3 kg) * (19.9 m/s)^2
Next, we can substitute the given values into the equation:
Work = (1/2) * (1.50 * 10^3 kg) * (19.9 m/s)^2 = 1.49 * 10^5 J
Now, we can calculate the average power developed by the engine:
Power = Work / Time = (1.49 * 10^5 J) / (11.6 s) ≈ 1 W (rounded to the nearest whole number)
Therefore, the average power developed by the engine is 1 W.
To find the instantaneous power output of the engine at t = 11.6 s (just before the car stops accelerating), we need to consider the forces acting on the car at that moment. At the given time, the car is still experiencing air resistance, which remains constant at 400 N throughout the acceleration period.
The instantaneous power output of the engine can be calculated using the equation:
Power = Force * Velocity
In this case, the force acting on the car is the sum of the engine force and the air resistance force:
Force = Engine Force + Air Resistance Force
The air resistance force is given as 400 N. The engine force can be found using Newton's second law, which states that Force = mass * acceleration. Since the car is accelerating uniformly, the engine force can be calculated as:
Engine Force = mass * acceleration
In this case, the mass is given as 1.50 multiplied by 103 kg and the acceleration can be calculated using the following equation:
Acceleration = (final velocity - initial velocity) / Time
Plugging in the given values, we get:
Acceleration = (19.9 m/s - 0 m/s) / 11.6 s ≈ 1.72 m/s^2 (rounded to two decimal places)
Now, we can calculate the engine force:
Engine Force = (1.50 * 10^3 kg) * (1.72 m/s^2) ≈ 2.58 * 10^3 N (rounded to two significant figures)
Finally, we can calculate the instantaneous power output of the engine:
Power = Force * Velocity = (2.58 * 10^3 N) * (19.9 m/s) ≈ 5.13 * 10^4 W (rounded to two significant figures)
Therefore, the instantaneous power output of the engine at t = 11.6 s is approximately 5.13 * 10^4 W.