Why Does Hooke's Law Seem Contradictory in Calculating Spring Energy?

Click For Summary
The discussion centers on the apparent contradiction between Hooke's Law, F=mg=kx, and the energy calculation, mg=(1/2)x*k. The confusion arises from the assumption that energy can be calculated using a constant force, which is not valid in this context. It is emphasized that the energy formula, E=force x displacement, only applies when the force remains constant throughout the displacement. The conversation suggests that understanding differential calculus may clarify the relationship between force and energy in spring systems. Proper measurement of the spring's displacement is crucial for accurate energy calculations.
tallwallyb
Messages
1
Reaction score
0
Homework Statement
Given a vertical spring with spring constant k=53N/m, a 2.5kg weight is hung from it. What is the change in x of the weight. (USE HOOKE'S LAW and compare it to using ENERGY.
Relevant Equations
F = kx
PEs = (1/2)kx^2
PEg = (mgh)
On the energy part, I keep getting mg=(1/2)*x*k, which is contradictory to Hooke's law F=mg=kx. What is going on?
 
Physics news on Phys.org
Please show how you applied energy considerations. Suppose you were given the spring and mass and you wanted to measure "the change in x of the weight." What procedure will you follow to do that? That procedure is important because it will inform your use of energy considerations.
 
tallwallyb said:
On the energy part, I keep getting mg=(1/2)*x*k, which is contradictory to Hooke's law F=mg=kx. What is going on?
I'm guessing you divided E by x expecting to get F. But the equation energy = force x displacement only works if the force is constant.
Are you familiar with differential calculus?
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
View full post »

Similar threads

Spring with mass hangs from the ceiling
Replies
22
Views
2K
Calculating the spring force constant K
Replies
14
Views
2K
Problem with when to use Force and Energy. What compresses spring/
Replies
3
Views
1K
Spring Problem Involving Variables and Constants Only
Replies
3
Views
1K
A block dropping onto a spring system
2
Replies
58
Views
2K
Vertical spring & maximum length
Replies
6
Views
1K
Initial velocity of bird landing on horizontal wire modeled as spring
Replies
7
Views
1K
Spring constant formlula for Force and Work Done
Replies
12
Views
2K
Difficulty in deciding when to apply work energy theorem
Replies
2
Views
1K
Time period of SHM & Energy conservation in pulley-spring-block system
Replies
24
Views
3K