Energy/work on incline plane with springs

[Panphobia]

ohhhhhhhhhhhh myyyyy, I didn't think of doing that, just adding the energies of m1 and m2. Now I get it so m1g(20+h-20sin40°) + m2g(40) + (1/2)k(20)2 = (1/2)m1v^2 + (1/2)m2v^2 + m2g20 + m1g(20 + h), then the m1g distributes and cancels with the mgh on the other side, and now it is solvable. Is that right?

 

[Tanya Sharma]

ohhhhhhhhhhhh myyyyy, I didn't think of doing that, just adding the energies of m1 and m2. Now I get it so m1g(20+h-20sin40°) + m2g(40) + (1/2)k(20)2 = (1/2)m1v^2 + (1/2)m2v^2 + m2g20 + m1g(20 + h), then the m1g distributes and cancels with the mgh on the other side, and now it is solvable. Is that right?

Excellent!

Well done

 

[Panphobia]

Man, you know what my problem is? I keep thinking back to high school physics where my teacher was awful, if I only used the knowledge I learned from the Uni lectures, I would be so much better off haha, but thanks :D

 

[Tanya Sharma]

m₁g(20+h-20sin40°) + m₂g(40) + (1/2)k(20)² = (1/2)(m₁+₂)v²+ m₂g(20) + m₁g(20+h)

Rearranging a bit , -m₁g(20sin40°) + m₂g(20) + (1/2)k(20)² = (1/2)(m₁+₂)v²

The above is what haruspex referrred in Post#27 .

I did not want to confuse you .First I wanted you to look at this approach and then thought of taking you to the better approach mentioned by haruspex .

Initially I had thought on the similar lines as haruspex ,but wasn't sure if that was simpler for you .

 

[Panphobia]

If my dumb self had thought of treating the objects and spring as one whole system, then everything would have made so much more sense, but thanks though, your assistance really helped, now I actually know the concept and when the exam comes I will actually know what to do in this type of situation.

 

[Tanya Sharma]

Good luck :thumbs: