The electric motor of a model train accelerates the train from rest to 0.680 m/s in 23.0 ms. The total mass of the train is 580 g. Find the average power delivered to the train during its acceleration.
power = work / time
work = force x distance
force = mass x acceleration
acceleration and distance can be calculated by....
Vf = Vo + at -----> a = (Vf -Vo) / t = Vf/t
and
d = Vot + 1/2 at^2 = 0 + 1/2 * (Vf/t) * t^2 = 1/2 * Vf * t
so....
power = mass x acceleration x distance / t
= m * a * d / t = m * (Vf/t) * (1/2 Vf t) /t = m Vf^2 / (2t)
= 580 g * (.664 m^2/s^2) / (2 * 17 * 10^-3 s) = 17.094 (kg m^2/s^2) / s = 17.094 Joule / sec = 17.094 watts
= 17.1 watts
To find the average power delivered to the train during its acceleration, we need to use the formula:
Power = Work/Time
First, let's find the work done on the train during its acceleration. The work done is equal to the change in kinetic energy of the train.
The formula for kinetic energy is:
Kinetic Energy = (1/2) * mass * velocity^2
Let's calculate the change in kinetic energy:
Initial kinetic energy = (1/2) * mass * initial velocity^2
Final kinetic energy = (1/2) * mass * final velocity^2
The change in kinetic energy is then:
ΔKE = Final kinetic energy - Initial kinetic energy
= (1/2) * mass * final velocity^2 - (1/2) * mass * initial velocity^2
= (1/2) * mass * (final velocity^2 - initial velocity^2)
Now, let's substitute the given values into the equation:
mass = 580 g = 0.58 kg
initial velocity = 0 m/s
final velocity = 0.680 m/s
ΔKE = (1/2) * 0.58 kg * (0.680 m/s)^2 - (1/2) * 0.58 kg * (0 m/s)^2
Simplifying the equation:
ΔKE = (1/2) * 0.58 kg * (0.680 m/s)^2
= 0.0588 J
Next, we need to calculate the time taken for the acceleration, given it is 23.0 ms. We need to convert milliseconds to seconds:
time = 23.0 ms = 23.0 * 10^(-3) s
Now, we can calculate the average power:
Power = Work / Time
= ΔKE / time
= 0.0588 J / (23.0 * 10^(-3) s)
≈ 2.55 W
Therefore, the average power delivered to the train during its acceleration is approximately 2.55 Watts.