A freight train has a mass of 1.4 107 kg. If the locomotive can exert a constant pull of 8.1 105 N, how long does it take to increase the speed of the train from rest to 66 km/h?
F = = ma,
a = F/m = 8.1*10^5 / 1.4*10^7 = 0.058m/s^2.
Vf = 66km/h * 1000m/km * (1/3600)h/s =
18.33m/s.
t = (Vf - Vo) / a,
t = (18.33 - 0) / 0.058 = 316s.
124.2 s
To find the time it takes for the train to increase its speed from rest to 66 km/h, we can use the concept of Newton's second law of motion.
Newton's second law states that the force acting on an object is equal to the product of its mass and acceleration, expressed as F = m * a.
In this case, we need to find the acceleration of the freight train. We can use the kinematic equation: v = u + at, where v is the final velocity, u is the initial velocity (which is 0 in this case since the train is at rest), a is the acceleration, and t is the time taken.
Converting the velocity from km/h to m/s, we have:
66 km/h = 66 * 1000 meters / 3600 seconds = 550 / 3 m/s
Using the formula v = u + at and plugging in the values:
550 / 3 = 0 + a * t
550t = 3a
Now, we need to find the acceleration of the train. Using Newton's second law, we know that F = m * a, where F is the force acting on the train and m is the mass of the train.
Given that the locomotive can exert a constant pull of 8.1 * 10^5 N and the mass of the train is 1.4 * 10^7 kg, we have:
8.1 * 10^5 N = 1.4 * 10^7 kg * a
a = (8.1 * 10^5 N) / (1.4 * 10^7 kg)
a ā 0.0579 m/sĀ²
Now we can substitute the value of acceleration into the equation 550t = 3a:
550t = 3 * 0.0579
t ā (3 * 0.0579) / 550
t ā 0.0317 seconds
Therefore, it takes approximately 0.0317 seconds to increase the speed of the train from rest to 66 km/h.