- #1
Tea_Aficionado
- 14
- 6
- Homework Statement
- A block of mass m = 2.0kg is dropped from height h = 40cm onto a spring of k = 1960N/m. Find the maximum distance the spring is compressed.
- Relevant Equations
- mgh = 1/2 * kx[SUB]max[/SUB][SUP]2[/SUP]
delta x = mg/k
h = x_max
We know that the Ug is converted to KE and Us. I thought that since the system loses energy after the collision that we shouldn't use the equation hnew= delta x + h.
I thought instead that maybe the h we should use is xmax, because that's when there is maximum Ug and there is no other energy present (having been converted all to Ug).
Therefore, mg * xmax= 1/2 * kxmax2
Solving for that with quad. eq, we get x = 0.02m
The solution, however, was 0.10m.
I looked at the worked-through examples online, and I'm confused as to why we don't consider the energy before and after the collision to be partially lost: their examples said to use hnew = h + Δx rather than just Δx.
If anyone could please take a bit of time to explain my thinking errors, I'd greatly appreciate it!
Post Notes:
I thought instead that maybe the h we should use is xmax, because that's when there is maximum Ug and there is no other energy present (having been converted all to Ug).
Therefore, mg * xmax= 1/2 * kxmax2
Solving for that with quad. eq, we get x = 0.02m
The solution, however, was 0.10m.
I looked at the worked-through examples online, and I'm confused as to why we don't consider the energy before and after the collision to be partially lost: their examples said to use hnew = h + Δx rather than just Δx.
If anyone could please take a bit of time to explain my thinking errors, I'd greatly appreciate it!
Post Notes:
- I don't know if I'm allowed to link the referenced examples here, but they seem for the most part agree with each other and can be found by copying and pasting the problem into the google search bar.
- I'm new, so if I made any formatting errors I apologize!