Coefficient of friction of a spring

Click For Summary
SUMMARY

The discussion centers on calculating the distance a spring was compressed and the velocity of a mass as it loses contact with the spring. A mass of 0.1676 kg is on a horizontal surface with a coefficient of kinetic friction (μk) of 0.340 and a spring with a force constant (k) of 508 N/m. The spring does 9.17 J of work on the mass, leading to the conclusion that the correct distance compressed is derived from the energy equation for springs, E = (1/2)kx², rather than using W = Fd without integration.

PREREQUISITES
  • Understanding of spring mechanics and Hooke's Law
  • Familiarity with work-energy principles
  • Knowledge of kinetic friction and its calculations
  • Ability to solve simultaneous equations in physics problems
NEXT STEPS
  • Study the derivation and application of the spring energy equation E = (1/2)kx²
  • Learn about the integration of forces in work calculations
  • Research the effects of friction on motion and energy loss
  • Explore advanced topics in dynamics involving mass-spring systems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of spring dynamics and energy conservation principles.

reaperkid
Messages
14
Reaction score
0

Homework Statement



A 1676 g mass is on a horizontal surface with mu-k = 0.340, and is in contact with a massless spring with a force constant of 508 N/m which is compressed. When the spring is released, it does 9.17 J of work on the mass while returning to its equilibrium position. Calculate the distance the spring was compressed. Then find the velocity of the mass as it loses contact with the spring?

m = .1676 kg
muk = .34
k = 508 N/m
w = 9.17 J

Homework Equations



W = Fd
x = F/k
muk = Fk / Fn

The Attempt at a Solution



I'm not sure if I'm even on the right track or not.
But I tried using simultanious equations and I got
d = squareroot (9.17/508) = .13 m but that wasn't right.
 
Physics news on Phys.org
Welcome to PF!

Hi reaperkid! Welcome to PF! :smile:

(have a mu: µ :wink:)
reaperkid said:
… a massless spring with a force constant of 508 N/m which is compressed. When the spring is released, it does 9.17 J of work on the mass while returning to its equilibrium position. Calculate the distance the spring was compressed.

Homework Equations


W = Fd
x = F/k
muk = Fk / Fn

… I tried using simultanious equations and I got
d = squareroot (9.17/508) = .13 m but that wasn't right.

You haven't written the standard energy equation for a spring …

E = (1/2)kx2 :smile:

(Your W = Fd gave you the wrong result, because you didn't integrate :wink:)
 

Similar threads

Replies
17
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
18
Views
3K
Replies
2
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
Replies
61
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
20
Views
3K