00:01
So here we can treat all 25 cars as a single object.
00:03
So the mass would be 25 multiplied by 5 .0 times 10 to the 24th to the 4th rather.
00:13
My apologies, kilograms.
00:16
And so we can substitute or rather convert 30 kilometers per hour, multiplied by 1 ,000 meters per kilometer, multiplied by one hour for every 3 ,600 seconds.
00:32
And this is giving us 8 .3 meters per second.
00:37
And so we know that this is subject to a frictional force f, lowercase f, is equaling, 25, multiplied by 250, multiplied by 8 .3.
00:50
And so this is giving us 5 .2 times 10 to the 4th newtons.
00:57
Now, at this point, we can say that for part a, along the level track, we can the object is experiencing a forward force t exerted by the locomotive.
01:09
So we can say the force of the train t minus the frictional force equaling the mass times the acceleration.
01:16
And so here the train force is equaling 5 .2 times 10 to the 4th newtons.
01:26
And this would be plus 1 .25 times 10 to the 6th kilograms, which is the mass...