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

[hbk69]

Tiny Tim and I discussed two different, but equivalent, methods of getting the same answer. Tim separated out the change in potential energy of the train from the work done by the engine and frictional force. In this approach, it is not necessary to include the component of the gravitational force acting along the grade. The potential energy change takes care of that.

In the approach that I discussed, I resolved the gravitational force into components perpendicular and parallel to the incline, and included the parallel component of weight explicitly in calculating the work done against external forces as the train moves along the incline. The work done against external forces is equal to the decrease in kinetic energy of the engine.

The method I gave and the method Tiny Tim gave are entirely equivalent, and lead to the exact same energy balance equation and the same distance traveled by the train.

Chet

PE? i thought it was ΔKE.

The energy method seems easier to grasp, although your method could probably be applied to most friction scenario? which means will need to understand it better

 

[haruspex]

As you are using the ΔKE are you not supposed to involve the final speed - the initial speed rather then Vo-V1? because in the question they say The train’s initial speed is 12 m/s. Through what horizontal distance will the train move before it’s speed is reduced to 9 m/s./QUOTE]
Final KE - initial KE gives you the gain in KE, which would of course be negative, so you'd write Final KE - initial KE = -7000d. Or you can write 7000d = loss of KE = initial - final.

 

[hbk69]

As you are using the ΔKE are you not supposed to involve the final speed - the initial speed rather then Vo-V1? because in the question they say The train’s initial speed is 12 m/s. Through what horizontal distance will the train move before it’s speed is reduced to 9 m/s./QUOTE]
Final KE - initial KE gives you the gain in KE, which would of course be negative, so you'd write Final KE - initial KE = -7000d. Or you can write 7000d = loss of KE = initial - final.

thanks understood, it was the math.

But i don't understand the way in which you got -7000, F = (3000-4000-6000) which formula are you using when you try and get the net force and why do you substract everything what is thetheory behind it?, also what is the 6000 value? where did it come from.

 

[haruspex]

But i don't understand the way in which you got -7000, F = (3000-4000-6000) which formula are you using when you try and get the net force and why do you substract everything what is thetheory behind it?, also what is the 6000 value? where did it come from.

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.

 

[Chestermiller]

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.

Very well said. I don't see how it could possibly be explained any better.

Chet

 

[hbk69]

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?

 

[haruspex]

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.)

 

[hbk69]

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

 

[haruspex]

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.

 

[hbk69]

thanks a million to everyone who has helped me