A freight train has a mass of 1.5*10^7 kg. If the locomotive can exert constant pull of 4.5 *10^5N, how long does it take to increase the speed of the train from res to 80.0km/hr?
a)467s b) 444s c)527s d) 425s e) 496s
To find the time it takes to increase the speed of the train, we can use Newton's second law, which states that the force exerted on an object is equal to the mass of the object multiplied by its acceleration.
First, let's convert the speed from kilometers per hour (km/hr) to meters per second (m/s).
1 km = 1000 m (since there are 1000 meters in a kilometer)
1 hour = 3600 seconds (since there are 3600 seconds in an hour)
So, 80.0 km/hr = (80.0 * 1000) / 3600 m/s ≈ 22.2 m/s.
Now, we can use the equation F = m * a, where F is the force, m is the mass, and a is the acceleration. In this case, the force F is given as 4.5 * 10^5 N and the mass m is given as 1.5 * 10^7 kg.
Rearranging the equation to solve for acceleration, we have:
a = F / m
Substituting the given values:
a = (4.5 * 10^5 N) / (1.5 * 10^7 kg) ≈ 0.03 m/s^2
Now, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity (0 m/s in this case), a is the acceleration, and t is the time. Rearranging this equation to solve for time, we have:
t = (v - u) / a
Substituting the given values:
t = (22.2 m/s - 0 m/s) / 0.03 m/s^2 ≈ 740 s
Therefore, it takes approximately 740 seconds to increase the speed of the train from rest to 80.0 km/hr.
Now let's check the options given:
a) 467s
b) 444s
c) 527s
d) 425s
e) 496s
The closest value to 740 seconds is 496 seconds, so the answer is (e) 496s.