Spring Energy Problem Conservative forces

AI Thread Summary
The discussion revolves around a physics problem involving a spring and a mass. The spring constant was calculated to be 87 N/m using the formula k=mg/x. The speed of the mass when it returns to the equilibrium length was determined to be approximately 0.94 m/s. For the lowest point of the spring after the mass is released, the extension was found to be about 0.18 m, leading to a total length of 0.43 m when combined with the relaxed length of the spring. The conversation emphasizes the importance of correctly defining energy states and relationships between height and spring extension.
GenericHbomb
Messages
3
Reaction score
0

Homework Statement


A spring is attached to a ceiling, and has a relaxed length of 25cm. When a mass m=.80kg is attached to the spring it stretches to an equilibrium length of L0=34cm.
a.)Find the Spring Constant of the Spring?

b.)I lift the mass until the spring returns to its relaxed length, and then release it. When the mass returns to the equilibrium length, what is its speed?

c.)After I release the mass and it falls, what is the length of the spring when the mass reaches its lowest point?

Homework Equations


K=mg/x

E1 + Wnc = E2

The Attempt at a Solution



A.) k=mg/x
k= [.80(9.8)]/.09
k=87 N/m

b.) E1= E2
0=.5mv2 + .5kx2 +mgh
0=.5(.8)v2 + .5(87)(.092) + .8(9.8)(-.09)
v=.9397 m/s

c.) I am not really sure where to go on C. I tried setting E1=E2 up as follows but my final answer was smaller than the equilibrium length so it can't be right.
E1=E2
.5mv2 + .5kx2 +mgh=.5mv2 + .5kx2 +mgh
.5(.8)(.93972)+.80(9.8)(34)= .5(87)x2
 
Physics news on Phys.org
GenericHbomb said:
c.) I am not really sure where to go on C. I tried setting E1=E2 up as follows but my final answer was smaller than the equilibrium length so it can't be right.
E1=E2
.5mv2 + .5kx2 +mgh=.5mv2 + .5kx2 +mgh
.5(.8)(.93972)+.80(9.8)(34)= .5(87)x2

Let us not take the E1 and E2 you have chosen
let E1 is the E1 in (b) ie when mass is in your hands

and let E2 be condition when mass is at lowest point

now try the question again
 
So if I set E1 to be when the spring is at the relaxed length then E1=0
E2 is lowest point so velocity would be zero since it stopped moving
E1=E2
0= .5mv2 + .5kx2 +mgh
0=.5(.80)(0)+ .5(87)(x)2+ (.8)(9.8)(h)
I know I need to solve for x because that is spring length but I am not sure what to put for h.
If I assume h to be 0 then my final answer would be 0.
 
x is the extension in string ...

h is height decreased

what can be relation b/w them
try by making a diagram
 
So I made a diagram and thank you for reminding that x is extension not length.
We are starting from x=0 and ending at x=?
Since h can be whatever I want I made h=0 where x=0
Therefore x=-h
So
0=.5(.80)(0)+.5(87)(x2)-.8(9.8)(x)
x=0, .18023
Length at lowest point = x + length at E1
Length= .25 + .18023
Length= .43 meters?
 
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