The horizontal distance the train will move before its speed reduces?

In summary: N forwards or 2. … 4000 N backwards?what is the total work done?what is the work done (total work done) ?what is the work done (Fg·s) ?what is the work done by those...1. … 3000 N forwards or 2. … 4000 N backwards?what is the total work done?what is the work done (total work done) ?The horizontal distance can be found by equating the work done by all external forces (Fg, Fengine, Ffriction) to the change in kinetic energy of the train. This gives the equation Fg*s + Fengine
  • #36
haruspex said:
You have 3000N attempting to accelerate the train, 4000N of resistance tending to slow it down, and the component of gravity parallel to the track is also slowing it down. The gravity component is Mg*1%, where M is the mass of the train. Taking g = 10N/kg gives the 6000 result.

thanks, i understand the math of it. But with the physics in order to get the net force you substract all the forces acting in various directions from the force which is accelerating the train?
 
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  • #37
hbk69 said:
thanks, i understand the math of it. But with the physics in order to get the net force you substract all the forces acting in various directions from the force which is accelerating the train?
Not 'in various directions'. You can only add and subtract magnitudes of forces that are acting parallel. These are all parallel to the slope of the track. When you stand still, gravity applies a force to you pushing you down, and the floor applies an equal and opposite force pushing you up. Subtracting the one from the other gives you 0, so no net force, so no acceleration. It's the same principle here, except that you do end up with a nonzero net force.
(You can add and subtract forces that are not parallel, but that's done using vector addition, which is different.)
 
  • #38
haruspex said:
Not 'in various directions'. You can only add and subtract magnitudes of forces that are acting parallel. These are all parallel to the slope of the track. When you stand still, gravity applies a force to you pushing you down, and the floor applies an equal and opposite force pushing you up. Subtracting the one from the other gives you 0, so no net force, so no acceleration. It's the same principle here, except that you do end up with a nonzero net force.
(You can add and subtract forces that are not parallel, but that's done using vector addition, which is different.)

How do we know if a force is a magnitude? in this case did you calculate the magnitude? if so how
 
  • #39
hbk69 said:
How do we know if a force is a magnitude?
A force is not a magnitude. There are two aspects to a force, its magnitude (how big it is) and its direction. Every force has a magnitude.
If two forces are in the same direction you can add the forces simply by adding their magnitudes. If they are in opposite directions you can subtract the magnitudes. If they not parallel then its more complicated.
in this case did you calculate the magnitude? if so how
We've already been through that. The three magnitudes were 3000, 4000 and 6000. The 3000 was in one direction, the 4000 and 6000 in the completely opposite direction. So we combine them by adding the two in the same direction and subtracting the one in the other direction.
 
  • #40
thanks a million to everyone who has helped me
 

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