Final answer:
The magnitude of the force exerted on the train during deceleration is calculated using Newton’s second law of motion. With an initial velocity of 4 m/s, a final velocity of 1 m/s, a mass of 800,000 kg, and a deceleration over 120 seconds, the force is found to be 20,000 Newtons.
Step-by-step explanation:
The student has asked for the magnitude of the force exerted on a passenger train when it decelerates from 4 m/s to 1 m/s over 120 seconds. To find the force, we use Newton’s second law, F = ma, where ‘F’ is the force, ‘m’ is the mass, and ‘a’ is the acceleration. First, we calculate the deceleration by using the formula for acceleration: a = (Δv) / t, where Δv is the change in velocity and t is the time taken for that change.
The initial velocity (vi) is 4 m/s, the final velocity (vf) is 1 m/s, thus the change in velocity (Δv) is -3 m/s (since the train is slowing down). The time (t) is 120 seconds. Plugging these values into the acceleration formula gives us: a = (Δv) / t = (-3 m/s) / 120 s = -0.025 m/s². The negative sign indicates the direction of acceleration is opposite to the direction of motion (deceleration).
Now that we have the acceleration, we can calculate the force. The mass (m) of the train is 800,000 kg. Therefore, F = ma = (800,000 kg) x (-0.025 m/s²) = -20,000 N. The negative sign here indicates that the force is in the opposite direction of the velocity - it is a braking force. The magnitude of the force is simply the absolute value, 20,000 N, since the question asks for magnitude which is always positive.
The magnitude of the force on the train when it slows down is 20,000 Newtons.