Answer :
Final answer:
To calculate the braking force on the train, first convert units and calculate the average deceleration using the change in velocity over time. Then, apply Newton's second law of motion to find the force. The average magnitude of the braking force on a 3000-ton train decelerating from 72 km/h to 18 km/h in 30 seconds is 1,500,000 N.
Explanation:
The question asks for the calculation of the average magnitude of the braking force exerted on a 3000-ton train as it slows down from 72 km/h to 18 km/h in 30 seconds.
First, convert the train's mass to kilograms (1 ton = 1000 kg), so the mass m = 3000 tons = 3,000,000 kg. Convert velocities from km/h to m/s: initial velocity u = 72 km/h = 20 m/s, final velocity v = 18 km/h = 5 m/s. The time t = 30 seconds.
The average acceleration a can be calculated using the formula a = (∂v)/(∂t) where ∂v (change in velocity) = v - u = 5 m/s - 20 m/s = -15 m/s, and ∂t (change in time) = 30 s. Therefore, a = -15 m/s² / 30 s = -0.5 m/s².
Finally, to find the braking force F, we use Newton's second law of motion, F = ma. So, F = 3,000,000 kg * -0.5 m/s² = -1,500,000 N. The negative sign indicates that the force is applied in the direction opposite to the train's motion, which is characteristic of a braking force.
Therefore, the average magnitude of the braking force on the train is 1,500,000 N.
