A freight train has a mass of 1.3 x10^7 kg. if the locomotive can exert a constant pull of 5.4x10^5 N, how long would it take to increase the speed of the train from rest to 71 km/h? Disregard friction. Answer in units of s.
To find the time it takes for the freight train to reach a speed of 71 km/h, we can use Newton's second law of motion and the equations of motion.
First, let's convert the speed from km/h to m/s since the units of mass and force are in kilograms (kg) and Newtons (N), respectively.
1 km/h is equal to (1000 m/1 km) * (1 h/3600 s) = 1/3.6 m/s
Therefore, the speed of the train is (71 km/h) * (1/3.6 m/s) = 19.72 m/s.
Newton's second law of motion states that force is equal to mass multiplied by acceleration:
Force = mass * acceleration
In this case, the acceleration can be calculated using one of the equations of motion:
v = u + at
where:
- v is the final velocity (19.72 m/s),
- u is the initial velocity (0 m/s),
- a is the acceleration, and
- t is the time.
Rearranging the equation, we can solve for the acceleration:
a = (v - u) / t
Substituting the known values, we get:
5.4 x 10^5 N = (1.3 x 10^7 kg) * ((19.72 m/s - 0 m/s) / t)
Simplifying the equation, we have:
5.4 x 10^5 N = (1.3 x 10^7 kg) * (19.72 m/s / t)
To find t, we can rearrange the equation as follows:
t = (1.3 x 10^7 kg) * (19.72 m/s) / (5.4 x 10^5 N)
Now, let's substitute the given values to calculate t:
t = (1.3 x 10^7 kg) * (19.72 m/s) / (5.4 x 10^5 N)
t ≈ 4752.96 s
Therefore, it would take approximately 4752.96 seconds (s) to increase the speed of the train from rest to 71 km/h.