Assuming 95.0% efficiency for the conversion of electrical power by the motor, what current must the 12.0-V batteries of a 845-kg electric car be able to supply to do the following?
(a) accelerate from rest to 25.0 m/s in 1.00 min
(b) climb a 200-m high hill in 2.00 min at a constant 25.0 m/s speed while exerting 440 N of force to overcome air resistance and friction
a. Work = Change in kinetic energy:
Work = 0.5M*V^2-0.5M*Vo^2, Vo = 0, Work = 0.5M*V^2 = 0.5*845*25^2 = 264,063 J.
Po = W/t = 264063/60 = 4401 J/s. = 4401 Watts = Power out.
Pi = 4401/0.95 = 4633 Watts = Power in.
Pi = 12*I = 4633, I = 4633/12 = 386 Amps.
To calculate the current required by the electric car's batteries, we will need to use the formula for power:
Power = Force * Velocity
a) To calculate the current required to accelerate the electric car from rest to 25.0 m/s in 1.00 minute, we need to find the average force applied during the acceleration. The work done to accelerate the car is given by:
Work = (1/2) * mass * velocity^2
Force * distance = (1/2) * mass * velocity^2
Since the car starts from rest, the distance is given by:
Distance = (1/2) * acceleration * time^2
Simplifying the equation, we have:
Force = (1/2) * mass * (velocity^2) / ((1/2) * acceleration * time^2)
Now, we can substitute the given values into the equation to find the force:
Mass = 845 kg
Velocity = 25.0 m/s
Acceleration = velocity / time = 25.0 m/s / (1.00 minute * 60 s/minute) = 0.4167 m/s^2
Time = 1.00 minute = 60 s
Force = (1/2) * 845 kg * (25.0 m/s)^2 / ((1/2) * 0.4167 m/s^2 * (60 s)^2)
Solving this equation, we find:
Force = 27603.12 N
Now, we can calculate the power required to generate this force at a velocity of 25.0 m/s:
Power = Force * Velocity = 27603.12 N * 25.0 m/s = 690,078 W
To find the current, we also need to consider the motor efficiency:
Efficiency = 95.0% = 0.95
Power = Efficiency * Input Power
Input Power = Power / Efficiency = 690,078 W / 0.95 = 726,136 W
Since the electric potential difference is given as 12.0 V, we can calculate the current using the formula:
Current = Power / Voltage = 726,136 W / 12.0 V = 60,511.33 A
Therefore, the current required for the electric car to accelerate from rest to 25.0 m/s in 1.00 minute is approximately 60,511.33 A.
b) To calculate the current required to climb a 200 m high hill in 2.00 minutes at a constant 25.0 m/s speed while exerting a force of 440 N to overcome air resistance and friction, we need to find the net force acting on the car.
The force required to overcome air resistance and friction can be calculated using the equation:
Force = Mass * Acceleration
Since the car is moving at a constant speed of 25.0 m/s, the net force is zero. Therefore, the force required to overcome air resistance and friction is equal in magnitude and opposite in direction to the force exerted by the car's motor.
Force = 440 N
Now, we can again use the equation:
Power = Force * Velocity
Power = 440 N * 25.0 m/s = 11,000 W
Considering the motor efficiency, we can calculate the input power:
Input Power = Power / Efficiency = 11,000 W / 0.95 = 11,578.95 W
To find the current, we can use the formula:
Current = Power / Voltage = 11,578.95 W / 12.0 V = 964.91 A
Therefore, the current required for the electric car to climb a 200 m high hill in 2.00 minutes at a constant 25.0 m/s speed while exerting a force of 440 N is approximately 964.91 A.
To solve this problem, we need to use the equations for power and electrical efficiency.
The power (P) supplied by the batteries can be calculated using the equation:
P = IV,
where I is the current and V is the voltage.
(a) To find the current required to accelerate the car, we can use the equation for power:
P = ΔE/Δt,
where ΔE is the change in energy and Δt is the time interval.
The change in energy can be calculated using the equation:
ΔE = 0.5mv^2,
where m is the mass of the car and v is the final velocity.
Plugging in the given values:
m = 845 kg,
v = 25.0 m/s,
ΔE = 0.5 * 845 kg * (25.0 m/s)^2 = 529,375 J.
Next, we need to determine the time interval in seconds. Since the time given is in minutes, we convert it to seconds:
Δt = 1.00 min * 60 s/min = 60 s.
Using the equation for power:
P = ΔE/Δt = 529,375 J / 60 s ≈ 8822 W.
The efficiency of the motor is given as 95.0%, so we need to calculate the input power (Pin) using the equation:
Pin = P/η,
where η is the efficiency as a decimal fraction (0.95 in this case).
Plugging in the given values:
Pin = 8822 W / 0.95 ≈ 9269 W.
Now, we can find the current (I) using the equation for power:
P = IV.
Rearranging the equation:
I = P/V,
where P is the power and V is the voltage.
Plugging in the given values:
I = 9269 W / 12.0 V ≈ 772 A.
Therefore, the batteries must be able to supply a current of approximately 772 A to accelerate the car.
(b) To find the current required to climb the hill while overcoming air resistance and friction, we need to calculate the power required first.
The power to overcome the force due to air resistance and friction can be calculated using the equation:
P = Fd/t,
where F is the force, d is the distance, and t is the time.
Plugging in the given values:
F = 440 N,
d = 200 m,
t = 2.00 min * 60 s/min = 120 s,
P = 440 N * 200 m / 120 s = 733 W.
Using the efficiency (η) of the motor and the equation for input power (Pin = P/η) as explained in part (a), we can find the current (I) needed.
Plugging in the given values:
Pin = 733 W / 0.95 ≈ 771 W.
Using the equation for current:
I = P/V,
where P is the power and V is the voltage, we can find the current (I).
Plugging in the given values:
I = 771 W / 12.0 V ≈ 64.3 A.
Therefore, the batteries must be able to supply a current of approximately 64.3 A to climb the hill at a constant speed of 25.0 m/s while overcoming air resistance and friction.