A train has a mass of 5.27E+6 kg and is moving at 53.9 km/hr. The engineer applies the brakes, which results in a net backward force of 1.30E+6 N on the train. The brakes are held on for 25.6 s. What is the new speed of the train? How far does it travel during this period?
just use you tried and true
v = v0 + at
s = vt + 1/2 at^2
a = F/m
I have tried the a = F/m, and it doesn't work. Also, the 2 top equations you wrote, I'm not too sure on how to use them.. I know the 53.9 km/hr is 14.97 m/s
what do you mean F/m "doesn't work"?
a = -1.30E6/5.27E6 = -.247
v = 14.97 - .247(25.6) = 8.65
s = 8.65*25.6 - .124*25.6^2 = 140.2
Oops. Got to use the initial velocity:
s = 14.97*25.6 - .124*25.6^2 = 302.0
To find the new speed of the train, we can use the equation:
Acceleration = Net Force / Mass
First, let's convert the train's initial speed from km/hr to m/s:
Initial Speed = 53.9 km/hr = (53.9 * 1000) / 3600 m/s = 14.9722 m/s
Given:
Mass of the train (m) = 5.27E+6 kg
Net backward force (F) = 1.30E+6 N
Now, using Newton's second law of motion, we can find the acceleration of the train:
Acceleration = Net Force / Mass
Acceleration = (1.30E+6 N) / (5.27E+6 kg)
Acceleration = 0.2466 m/s²
Since the brakes are held on for 25.6 seconds, we can use the kinematic equation:
Final Speed = Initial Speed + (Acceleration * Time)
Final Speed = 14.9722 m/s + (0.2466 m/s² * 25.6 s)
Final Speed = 20.2947 m/s
Therefore, the new speed of the train is approximately 20.29 m/s.
To find the distance traveled during this period, we can use another kinematic equation:
Distance = Initial Speed * Time + (0.5 * Acceleration * Time²)
Distance = 14.9722 m/s * 25.6 s + (0.5 * 0.2466 m/s² * (25.6 s)²)
Distance = 383.2202 m + (0.5 * 0.2466 m/s² * 655.36 s²)
Distance = 383.2202 m + 80.87121216 m
Distance = 464.09141216 m
Therefore, the train travels approximately 464.09 meters during this period.