A freight train has mass of 1.6 x 107 kg. If the locomotive can exert a constant pull of 760000 N, how long does it take to increase the speed of the train from rest to 79 km/h? Answer in units of min.
To find the time it takes to increase the speed of the train, we can use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass.
Here's how we can solve this problem step by step:
Step 1: Convert the speed of the train from km/h to m/s.
We know that 1 km/h is equal to 1000 m/3600 s. So, 79 km/h is equal to (79 x 1000) / 3600 m/s.
79 km/h = (79 x 1000) / 3600 m/s = 79000/3600 m/s ≈ 21.94 m/s
Step 2: Calculate the acceleration of the train.
We can use the formula v = u + at, where:
- v is the final velocity (21.94 m/s),
- u is the initial velocity (0 m/s),
- a represents the acceleration (which we need to find), and
- t is the time taken.
Since the initial velocity is 0, the equation simplifies to v = at.
Rearranging the equation, we have a = v / t.
Step 3: Find the acceleration of the train.
Since the force applied (F) is equal to the mass (m) multiplied by the acceleration (a), we can rearrange the equation to solve for a:
F = ma
a = F / m
Plugging in the values we know:
F = 760,000 N (force applied by the locomotive)
m = 1.6 x 10^7 kg (mass of the train)
a = (760,000 N) / (1.6 x 10^7 kg) ≈ 0.0475 m/s^2
Step 4: Substitute the acceleration into the equation v = at to find the time, t.
Using the rearranged equation, we can solve for t:
t = v / a
Substituting the known values:
t = 21.94 m/s / 0.0475 m/s^2 ≈ 461.25 s
Step 5: Convert the time from seconds to minutes.
To convert seconds to minutes, we divide the time by 60.
t(minutes) = 461.25 s / 60 ≈ 7.688 minutes
Therefore, it takes approximately 7.688 minutes to increase the speed of the train from rest to 79 km/h.