Multiple concept self-made passage at 1:30 AM. Questions ahoy.

[pfaction]

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So I couldn't sleep and made my own passage and questions up. I just want to make sure the things are right.

You have an incline that is 5 meters tall and 10 meters long, angle of 30 degrees. A 10kg block is on the frictionless plane, at rest, and is allowed to slide down. At the bottom, 1 meter of a frictionless surface separates the base and a spring with a constant of 10N/m. The spring recoils and shoots the 10kg block back up the incline. How far from the start point does the block reach?

So for this, I had the initial questions: Is the final speed and/or the final time the same when sliding down a frictionless incline as it is being dropped straight down, as in free fall?

So as we all know, v = sqrt (2 g h) so 20*5; the speed would be 10 if I dropped it straight. For the incline, it would be vsq = v0sq + 2ad; where a is gsinO and d is 10, so it would be the same thing i think: sqrt (2*10sin30*10*) is 10.
So the speed is 10 m/second either way, but I think the times would differ. 5=5tsq for free fall, 10=(.5)(10sin30)(tsq) so it'd be 4 = tsq I think: the incline would take twice as long as free fall.

So then it slides one meter and hits the spring. Now this is where it starts to get confusing for me. Should I use F=mg =-kx? Or should I use energy; PE=mgh=KE = 0.5(k)(xsq)? If I use F=mg=kx; 10(10) = 10(x); the spring is compressed 10 meters on recoil. If I use PE; (100)(5) = (5)(xsq) and it's recoiled 10 meters. I also thought about using impulse but there's no time given, and work is FdeltaV...so again, not sure if I can use that here.

I'm going to insert a side question: If a 10kg object at 10m/s and 20kg object at 5m/s speed slam into a wall, which has the larger force?

In either case, the spring recoils 10 meters and shoots the 10kg back up the 10 meter incline. Since the energy was conserved, the speed didn't change, it still has 500J of energy. But this time, gravity is working against the object. W=Fd; 500 = (mgsinO)(d) = (10*5)(d) = 10 meters? It would go directly back up the incline?

👍

 

[EnginrTheFuture]

So I couldn't sleep and made my own passage and questions up. I just want to make sure the things are right.

You have an incline that is 5 meters tall and 10 meters long, angle of 30 degrees. A 10kg block is on the frictionless plane, at rest, and is allowed to slide down. At the bottom, 1 meter of a frictionless surface separates the base and a spring with a constant of 10N/m. The spring recoils and shoots the 10kg block back up the incline. How far from the start point does the block reach?

1) So for this, I had the initial questions: Is the final speed and/or the final time the same when sliding down a frictionless incline as it is being dropped straight down, as in free fall?

So as we all know, v = sqrt (2 g h) so 20*5; the speed would be 10 if I dropped it straight. For the incline, it would be vsq = v0sq + 2ad; where a is gsinO and d is 10, so it would be the same thing i think: sqrt (2*10sin30*10*) is 10.
So the speed is 10 m/second either way, but I think the times would differ. 5=5tsq for free fall, 10=(.5)(10sin30)(tsq) so it'd be 4 = tsq I think: the incline would take twice as long as free fall.

So then it slides one meter and hits the spring. Now this is where it starts to get confusing for me. Should I use F=mg =-kx? Or should I use energy; PE=mgh=KE = 0.5(k)(xsq)? If I use F=mg=kx; 10(10) = 10(x); the spring is compressed 10 meters on recoil. If I use PE; (100)(5) = (5)(xsq) and it's recoiled 10 meters. I also thought about using impulse but there's no time given, and work is FdeltaV...so again, not sure if I can use that here.

I'm going to insert a side question: If a 10kg object at 10m/s and 20kg object at 5m/s speed slam into a wall, which has the larger force?

In either case, the spring recoils 10 meters and shoots the 10kg back up the 10 meter incline. Since the energy was conserved, the speed didn't change, it still has 500J of energy. But this time, gravity is working against the object. W=Fd; 500 = (mgsinO)(d) = (10*5)(d) = 10 meters? It would go directly back up the incline?

👍

1) No for time. Final speed is same due to conservation of energy, but that's not direction specific. By putting it on an inclined plane, the acceleration in the y direction is impeded by the plane. Essentially you can interpret this as converting your potential energy into x direction kinetic energy rather than y direction. If your plane is straight up and down (free fall), your object takes the potential energy and converts it directly into KE downward (y direction). This is why with free fall, it reaches the ground faster, because the average y velocity of the trip is so much higher. As you make the plane more and more flat, more and more PE goes into x direction KE. As you know, the amount of PE is the same as the final total energy. So, you know the final y direction KE (velocity in y direction) cannot be the same as the free fall because you have x direction KE that had to come from somewhere. Thus the final Y velocity is slower on the plane, and it takes more time to get down.

If you want to visualize this on the free body diagram. Draw the plane and the mass. The only movement is in the direction of the force parallel to the plane. I assume you broke down the Weight force into the two standard component forces parallel and perpendicular to the plane. That parallel force has two components itself. A directly down and directly sideways one. This is the only force that causes motion and so these components tell you the actual x and y accelerations. As you can see, as the plane gets flat, both components get really small. The block won't move just sitting on a table for this reason. The movement perpendicular to the plane is ignored because the block can't traverse through the surface just like in the above example. The parallel force in both its downward and sideways components becomes 0. Block doesn't move.

 

[chiddler]

Only answering the initial question, without reading the !!massive!! amounts of text you guys wrote:

It would start and end at the same position because of conservation of energy. Think about how much energy it starts with. If no friction, the energy isn't going anywhere. So it goes right back to where it started.