Christopher B. answered • 02/25/22
Tutor
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STEM Tutor/Teacher Specializing in Mechanical Physics
Hey Tapti,
Hope I'm interpreting this correctly - sounds like the spring is underneath the 5-kg block. So, here's what I would do:
- Use the ∆x1 value and the total mass of the two blocks to calculate the spring constant, k, using Hooke's law.
- Use Hooke's law again to determine where the 5-kg block will find equilibrium after the 3kg block is lifted off of it.
- Now you have a conservation of energy problem
- In the final position, we can say that the blocks have a GPE of 0, when the spring is compressed to its maximum at the bottom of the oscillation (maybe call this ∆xfinal). At this point in the oscillation, the blocks will also have a KE of 0, so all of the energy in the system will be in the form of Elastic PE.
- For the initial energy of the system, you'll have to take into account two things:
- The 3 kg block has a GPE associated with its height of 0.750 m above the equilibrium position of the other block. This term you can actually put a number to (in Joules) right away.
- BOTH blocks will then lose more GPE as they descend together from that position to the bottom of the oscillation. This term will be associated with the variable ∆xfinal
It looks like you may have a quadratic equation on your hands once you get this all set up. Good luck!
Christopher B.
tutor
Oh I'm sorry! I missed that it was an inelastic collision. Looks like you have K correct as well as the compression of the spring under the 5-kg block alone. The height of the 5-kg block shouldn't matter though, as it says that the 3-kg block is just lifted .750 m above that block.
So yes, once you determine how fast the 3-kg block is moving after falling that distance, you should use conservation of momentum to find the speed of the blocks (which are now 8kg together) when the spring is at that position of x = .125m. I have v = 1.452m/s at this point.
Then, yes, you would use energy, and I see where you're stuck. The h in your "mgh" term is really x - .125m, since we've already accounted for any PE loss associated with moving from the true equilibrium position down to x = .125m. So this must be what your teacher meant by the system relating x to h.
So that gives you: 1/2k(.125)^2 + 1/2mv^2 + mg(x - .125) = 1/2kx^2.
Once you plug in what you know, NOW you'll have a quadratic on your hands, and I imagine only one of the solutions will make sense.
I hope this helps and sorry if I caused any confusion.
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02/28/22
Tapti C.
thank you!
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03/01/22
Tapti C.
okay so i determined the correct value of K was 400 N/m. i have been using the pointers you gave and i still don't get it. the thing is, height and the x value (compression) are treated differently in this problem, so my teacher said to create a system to relate h to x. so in essence, after using conservation of momentum to find the v final of the 8kg block, you would use an energetics equation to do 1/2kx^2 +1/2mv^2 +mgh =1/2kx^2 (bottom) but if you don't know h, you cant use this equation. i tried using energetics to relate x to h, but that's wrong, and i am stuck at this point. The 5kg block is above the spring, and the spring is initially uncompressed (x=0) and then the 5kg block compresses it by 0.125 m. I don't know the height of the 5kg block to calculate mgh.02/27/22