A freight train has a mass of 1.2 × 107 kg. If the locomotive can exert a constant pull of 5.0 × 105 N, how long would it take to increase the speed of the train from rest to
89.2 km/h? (Disregard friction.) Answer in units of s.
The acceleration rate with that force acting is
a = F/m = 0.0417 m/s^2
89.2 km/h = 24.78 m/s is the speed you with to attain.
Solve this equation for the required time, t:
(1/2) a t^2 = 24.78 m/s
To solve this problem, we need to use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
In this case, the net force is the force exerted by the locomotive, the mass is the mass of the freight train, and the acceleration is the change in velocity over time.
First, let's convert the final speed from km/h to m/s since SI units are used in Newton's second law of motion:
89.2 km/h * (1000 m/1 km) * (1 h/3600 s) = 24.78 m/s
Next, we can use the equation for acceleration:
acceleration = change in velocity / time
Since the train starts from rest, the change in velocity is just the final velocity:
acceleration = 24.78 m/s / time
Now, let's rearrange the equation to solve for time:
time = 24.78 m/s / acceleration
To find the acceleration, we can use Newton's second law:
acceleration = net force / mass
In this case, the net force is the force exerted by the locomotive, and the mass is the mass of the freight train.
acceleration = (5.0 × 10^5 N) / (1.2 × 10^7 kg)
Now, let's substitute this value back into the equation for time:
time = 24.78 m/s / [(5.0 × 10^5 N) / (1.2 × 10^7 kg)]
Simplifying this expression:
time = (24.78 m/s) * [(1.2 × 10^7 kg) / (5.0 × 10^5 N)]
Calculating the multiplication:
time = 595.2 s
Therefore, it would take approximately 595.2 seconds to increase the speed of the train from rest to 89.2 km/h, disregarding friction.