Very simple spring constant questoin

AI Thread Summary
The spring constant for a vertical spring is defined as k=mg/Δ, while for a horizontal spring on a frictionless surface, it follows the equation F = kx. In horizontal motion, the acceleration 'a' in F=ma corresponds to the acceleration of the mass, not gravitational acceleration 'g'. It is crucial to remember that Newton's second law states that the sum of all forces equals mass times acceleration, not just a single force. Misunderstanding the relationship between forces in these equations can lead to confusion and errors in physics.
The thinker
Messages
54
Reaction score
0
Hi,

I've just been thinking about this too much and managed to confuse myself.

Spring constant for a vertical spring: k=mg/\Delta

Is it the same for a horizontal spring on a frictionless surface?

Thanks!
 
Physics news on Phys.org
No. Horizontally, it's just F = kx.
 
Limecat said:
No. Horizontally, it's just F = kx.

Thanks

So the "a" in F=ma in the horizontal case is just whatever acceleration the mass is undergoing, rather than g?
 
The thinker said:
Thanks

So the "a" in F=ma in the horizontal case is just whatever acceleration the mass is undergoing, rather than g?

Yes,you are right.
During the horizontal motion

F= ma = kx

And during the vertical motion

F = mg = kx

If friction is taken into count that force also shall be taken into count.
 
Newton's 2nd law says
Sum of Forces = m*a
not just F = m*a

When written as F = ma, this tends to neglect that it is the sum of all forces on the particle (Newton's 2nd law is for a particle).

Writing F = mg is prone to suggest errors. It is safer to write
W = m*g
as a reminder that it is the weight force specifically that is equal to mg.

I am continually amazed at the number of people I see posting here who want to confuse the F's in F = ma and F = mg by setting them equal to each other! There is a dangerous lack of thinking going on.
 
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 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
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