A 2300. kg car starts from rest and accelerates uniformly to 24.0 m/s in 24.0 s. Assume that air resistance remains constant at 510 N during this time. Find (a) the net force on the car
b) the average power developed by the engine?
c) the instantaneous power output of the engine at t = 24.0 s, just before the car stops accelerating.
To find the answers to the given questions, we will first need to calculate the net force experienced by the car throughout the acceleration, and then we can use this information to find the average power developed by the engine. Lastly, we can use the instantaneous power formula to find the power output of the engine at t = 24.0 s.
(a) To calculate the net force on the car, we need to consider the forces acting on it. In this case, the only force mentioned is air resistance, which remains constant at 510 N. So, the net force can be calculated using Newton's second law, F_net = ma, where F_net is the net force, m is the mass of the car, and a is the acceleration.
Given:
Mass of the car (m) = 2300 kg
Final velocity (v) = 24.0 m/s
Time (t) = 24.0 s
Air resistance (F_resistance) = 510 N
Using the formula for acceleration (a):
a = (v - u) / t
where u is the initial velocity. Here, the car starts from rest, so u = 0 m/s.
Substituting the values, we can find the acceleration:
a = (24.0 m/s - 0 m/s) / 24.0 s
a = 1.0 m/s^2
Now, we can calculate the net force:
F_net = ma
F_net = (2300 kg)(1.0 m/s^2)
F_net = 2300 N
Therefore, the net force acting on the car is 2300 N.
(b) To calculate the average power developed by the engine, we can use the work-energy principle. The work done by the net force during the acceleration is equal to the change in kinetic energy.
The work done (W) is given by the formula:
W = F_net * d
where d is the displacement of the car during acceleration.
The displacement can be calculated using the formula:
d = ut + (1/2)at^2
where u is the initial velocity, t is the time, and a is the acceleration.
Substituting the values:
u = 0 m/s
t = 24.0 s
a = 1.0 m/s^2
d = (0 m/s)(24.0 s) + (1/2)(1.0 m/s^2)(24.0 s)^2
d = 288 m
Now, calculating the work done:
W = F_net * d
W = (2300 N)(288 m)
W ≈ 662,400 J (Joules)
The average power developed by the engine is given by the formula:
P_avg = W / t
where P_avg is the average power and t is the time taken.
Substituting the values:
P_avg = 662,400 J / 24.0 s
P_avg ≈ 27,600 W (Watts)
Therefore, the average power developed by the engine is approximately 27,600 Watts.
(c) To find the instantaneous power output of the engine at t = 24.0 s, just before the car stops accelerating, we can use the formula:
P = F_net * v
where P is the power, F_net is the net force, and v is the velocity.
Substituting the values:
P = (2300 N)(24.0 m/s)
P ≈ 55,200 W (Watts)
Therefore, the instantaneous power output of the engine at t = 24.0 s is approximately 55,200 Watts.