Question

Problem regarding conservation of energy (see attachment)

Answer 1

My work thus far:
\[\large T_A+V_A=T_B+V_B\]
\[\large T_A=0\]\[\large V_A=0.5k(\Delta l)=0.5(37)(\color{red}{????})\]\[\large T_B=0.5mv_b^2=0.5(1)v_b^2\]\[\large V_B=0.5k(\Delta l)+W*h\]

Where T is the kinetic energy and V is the sum of the grav/spring potential energies.

I think my problem at the moment, is determining what goes inside the "?????"
I can't seem to determine the stretched length of the spring at point A. I think once I have that piece of information, I should be able to solve the rest.

Answer 2

Nice!
Look , you might be able to catch this up!
1/2k(x1^2-x2^2)=1/2mv^2
x1,x2 are elongation in spring at A and B positions respectively.

Answer 3

@samigupta8 Yeah, but my issue is determining x1, or in my post, \(\large \Delta l_A\)

Answer 4

@agent0smith =]

Answer 5

It's easy to determine the spring elongation at point B... notice you can draw a right triangle with the hypotenuse as the spring

Answer 6

I'd rather just do it this way\[\large \frac{ 1 }{ 2 }mv_A^2 + mgh_A + \frac{ 1 }{2 }kx_A^2=\frac{ 1 }{ 2 }mv_B^2 + mgh_B + \frac{ 1 }{2 }kx_B^2\]
But you can make things like point A the zero point for grav. PE, and it's at rest at A.\[\large \frac{ 1 }{2 }kx_A^2=\frac{ 1 }{ 2 }mv_B^2 + mgh_B + \frac{ 1 }{2 }kx_B^2\]
I just realized the units are unbelievably not in metric units. I'm blown away by that. For that reason, I am not going to list the units, because they are stupid, and this problem has now become stupid by extension. What a shame, it was kinda interesting.

I won't list any numbers (because... stupid). But you should be able to figure out the spring stretch at A and B, given the info in the question and what I said about a right triangle (don't forget the undeformed length!). Height at B is below the zero point so it's negative.

Answer 7

For the normal force at point B, I'm assuming it's upwards

Answer 8

Using the right triangle I mentioned earlier, you can find the angle, and thus find the components of the spring force.

Keep in mind that at point B it's in circular motion, so the net force would be mv^2/r. Make positive upward. You have Fn and the vert. comp. of F spring acting upward (Fys below), and mg downward \[\large \frac{ mv_B^2 }{ r }=F_n+F_{ys}-mg\]

Answer 9

See the diagram, study the given numbers in it, to help figure out the spring stretch at A and B (it's a right triangle). Don't forget you only want the stretch past the natural length...

Answer 10

I think i get it now :S I was just having trouble determining the stretched length of the spring at point A, is all