A freight train has a mass of 1.5 107 kg. If the locomotive can exert a constant pull of 6.9 105 N, how long does it take to increase the speed of the train from rest to 72 km/h?
min
0.98
i assume you mean the mass is
M = 1.5*10^7 kg
The acceleration is
a = F/M = 6.9*10^5/1.5*10^7 = 0.046 m/s^2
72 km/h = 20 m/s
Divide 20 m/s by the acceleration rate to get the time required.
T = (20 m/s)/(0.046 m/s^2) = _____ s
Well, my friend, let's chug along with this calculation!
First, let's convert the speed from km/h to m/s. To do that, we need to multiply 72 km/h by 1000 (to convert it to meters) and then divide it by 3600 (to convert it to seconds). This gives us a speed of 20 m/s.
Now, we need to calculate the acceleration of the train. We can use the basic equation: force = mass × acceleration. Rearranging the equation to solve for acceleration gives us acceleration = force / mass. Plugging in the given values, we have acceleration = 6.9 × 10^5 N / 1.5 × 10^7 kg, which simplifies to 0.046 m/s^2.
To find the time it takes for the train to reach a speed of 20 m/s, we can use the equation: final speed = initial speed + acceleration × time. Rearranging the equation to solve for time gives us time = (final speed - initial speed) / acceleration. Plugging in the values, we have time = (20 m/s - 0 m/s) / 0.046 m/s^2. Crunching the numbers, this gives us a time of approximately 435 seconds.
So, my friend, it will take roughly 435 seconds for the train to reach a speed of 72 km/h. Just enough time for the train conductor to grab a cup of coffee and enjoy the ride!
To find the time it takes for the freight train to increase its speed from rest to 72 km/h, we can use the concept of Newton's second law of motion.
First, let's convert the speed of the train from kilometers per hour (km/h) to meters per second (m/s) since the unit of force is in Newtons (N), and the unit of mass is in kilograms (kg).
Speed in m/s = (72 km/h) * (1000 m/km) * (1 h/3600 s) ≈ 20 m/s
Now, using Newton's second law of motion, we have:
Force (F) = Mass (m) * Acceleration (a)
In this case, the force is the constant pull provided by the locomotive. The mass is the mass of the train, and we need to find the acceleration.
Rearranging the formula to solve for acceleration:
Acceleration (a) = Force (F) / Mass (m)
Substituting the given values:
Acceleration (a) = 6.9 * 10^5 N / 1.5 * 10^7 kg ≈ 0.046 m/s^2
Now, we can use the kinematic equation:
Final velocity (v) = Initial velocity (u) + Acceleration (a) * Time (t)
Since the initial velocity (u) is zero (as the train is at rest initially), we can simplify the equation to:
Final velocity (v) = Acceleration (a) * Time (t)
Rearranging the equation to solve for time:
Time (t) = Final velocity (v) / Acceleration (a)
Substituting the given values:
Time (t) = 20 m/s / 0.046 m/s^2 ≈ 435.9 s (approximately)
Therefore, it takes approximately 435.9 seconds (or 7 minutes and 15.9 seconds) to increase the speed of the train from rest to 72 km/h.