How much total kinetic energy is available in a diesel locomotive weighing 11000 kg travelling at 2.0 km/h?
K.E. = (1/2) M V^2
is the answer 28.51 mJ
28.51 milliJoules? I don't think so. That's much too low
When you use the formula, V must be in m/s.
2.0 km/h = 0.5556 m/s
To calculate the total kinetic energy of the diesel locomotive, we need to use the formula:
Kinetic Energy = (1/2) * mass * velocity^2
First, we need to convert the mass of the locomotive from kilograms to grams since the formula uses grams as the unit for mass. We can do this by multiplying the mass by 1000:
11000 kg = 11000 * 1000 g = 11,000,000 g
Next, we need to convert the velocity from kilometers per hour (km/h) to meters per second (m/s) since the formula uses meters per second. We can do this by following these steps:
Step 1: Convert km/h to m/min by multiplying the velocity by 1000 (since there are 1000 meters in a kilometer):
2.0 km/h = 2.0 * 1000 m/min = 2000 m/min
Step 2: Convert m/min to m/s by dividing the velocity by 60 (since there are 60 seconds in a minute):
2000 m/min = 2000 / 60 m/s = 33.33 m/s (rounded to two decimal places)
Now, we have all the required values to calculate the kinetic energy. Let's plug them into the formula:
Kinetic Energy = (1/2) * mass * velocity^2
= (1/2) * 11,000,000 g * (33.33 m/s)^2
Calculating this, we get:
Kinetic Energy ≈ (1/2) * 11,000,000 g * (1115.5889 m^2/s^2)
≈ 6,119,723,000 g * m^2/s^2
Note that the unit for kinetic energy in the International System of Units (SI) is Joules (J). Therefore, we need to convert the unit from grams multiplied by square meters per second squared (g * m^2/s^2) to Joules. To do this, we divide the calculated value by 1000:
6,119,723,000 g * m^2/s^2 = 6,119,723,000 g * m^2/s^2 / 1000 = 6,119,723 J
Therefore, the total kinetic energy available in the diesel locomotive is approximately 6,119,723 Joules.