Max Velocity While Compressing a Spring?

AI Thread Summary
The discussion revolves around a physics problem involving a 17 kg box sliding down a frictionless ramp and colliding with a spring with a spring constant of 210 N/m. The first part of the problem, concerning the maximum compression of the spring, has been solved correctly at 2.2 m. The second part, which asks for the compression at which the box has its maximum speed, is more contentious, with some participants mistakenly suggesting that the answer is zero compression. The correct approach involves understanding that the box continues to accelerate until the forces balance, specifically when mgsin(theta) equals kx. The user ultimately realizes a calculation error related to the mass, indicating the need for careful attention to detail in solving such problems.
peterbishop
Messages
4
Reaction score
0

Homework Statement



A 17 kg box slides 4.0 m down the frictionless ramp shown in the figure, then collides with a spring whose spring constant is 210 N/m. (picture attached)

a) What is the maximum compression of the spring? (solved this part)
b) At what compression of the spring does the box have its maximum speed?


Homework Equations



F = -kx
Kspring = 1/2 * k* x^2


The Attempt at a Solution



I solved part a already and the answer was 2.2 m, so that's all good. Part b I'm having some trouble with. I've seen this problem attempt to be answered on other threads but no one ever discusses part b thoroughly, it's usually about part a, which I didn't have a problem with. I've seen quite a few people on the internet say that the answer is zero compression, because the spring will slow the box down when it hits it, but this is not the case, as the spring only lessens the acceleration: it is still accelerating, I believe until a point when mgsin(theta) = kx, x being the compression of the spring. Using this I get an answer of .023 m though, which isn't correct either. Any help would be appreciated. (Also I tried putting zero for the answer just for the heck of it but masteringphysics says it's wrong as well.)
 

Attachments

  • physics.jpg
    physics.jpg
    3.2 KB · Views: 488
Physics news on Phys.org
peterbishop said:
I've seen quite a few people on the internet say that the answer is zero compression, because the spring will slow the box down when it hits it, but this is not the case, as the spring only lessens the acceleration: it is still accelerating, I believe until a point when mgsin(theta) = kx, x being the compression of the spring.
Sounds good to me.
Using this I get an answer of .023 m though, which isn't correct either.
How did you get that answer? Redo your calculation.
 
Yep you're right, thanks, for some reason whenever I wrote it down in my notes before I did the calculation I left out mass.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top