The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on th

Question

kondor19780726

28 days ago

13

The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on th

e locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 km/h?

Physics

1 answer:

Keith_Richards [3.2K] · Answer 1 · 2026-04-12T00:23:40+03:00

Answer:

It would take the freight train 542.265 seconds to accelerate from zero to 80.0 kilometers per hour.

Explanation:

The statement lacks completeness. The full explanation follows:

The mass of the freight train is $1.83\times 10^{7}\,kg$ . With a consistent net force of $7.50\times 10^{5}\,N$ exerted by the locomotive's wheels against the tracks, the same amount of force is applied forward onto the locomotive from the tracks. For this scenario, ignore aerodynamics and any other friction affecting the additional wheels of the train. How much time, in seconds, is required to boost the train's speed from a standstill to 80.0 kilometers per hour?

If the locomotive experiences a steady net force ( $F$ ), represented in newtons, then its acceleration ( $a$ ), expressed in meters per square second, remains constant and can be calculated using the equation:

$a = \frac{F}{m}$ (1)

Where $m$ denotes the mass of the freight train in kilograms.

If we are aware of $F = 7.50\times 10^{5}\,N$ and $m = 1.83\times 10^{7}\,kg$ , this leads to the train's acceleration being:

$a = \frac{7.50\times 10^{5}\,N}{1.83\times 10^{7}\,kg}$

$a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$

Now to identify the time for the freight train to accelerate from a complete stop ( $t$ ), measured in seconds, we apply the following formula:

$t = \frac{v-v_{o}}{a}$ (2)

Where:

$v$ - The concluded speed of the train in meters per second.

$v_{o}$ - The starting speed of the train in meters per second.

If we have values for $a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$ , $v_{o} = 0\,\frac{m}{s}$ , and $v = 22.222\,\frac{m}{s}$ , the time needed by the freight train comes out to be:

$t = \frac{22.222\,\frac{m}{s}-0\,\frac{m}{s} }{4.098\times 10^{-2}\,\frac{m}{s^{2}} }$

$t = 542.265\,s$

The freight train would take 542.265 seconds to increase the speed of the train from rest to 80.0 kilometers per hour.