Understanding Hooke's Law: Sample Problems and Solutions

  • Thread starter Thread starter meikamae
  • Start date Start date
  • Tags Tags
    Hooke's law Law
Click For Summary
Hooke's Law describes the relationship between the force exerted on a spring and its displacement from equilibrium, represented by the equation F = -kx, where F is the force, k is the spring constant, and x is the displacement. The law states that the force opposing the stretching of the spring increases linearly with distance. Sample problems can involve calculating the stretch of a spring given a force, determining the spring constant, or finding the necessary force for a specific displacement. The discussion emphasizes understanding the variables and their meanings in practical applications of Hooke's Law. Mastery of these concepts allows for deeper exploration into harmonic motion and related physics principles.
meikamae
I need a sample problem (and solution, please) for hooke's law to help me understand. I understand the equation, I just don't understand what the variables mean exactly and how the equation works and what each variable stands for.
 
Physics news on Phys.org
ok, Hooke's law applies to the idealized case of a spring. The further you stretch the spring, the greater the force opposing the stretching, in other words, it assumes that the force increases linearly with distance.
F = -kx
where k is the spring constant, F is the force generated by the spring, x is the displacement from equilibrium (where F=0). Any basic sample problem will require the equation re-arranged; or substitution of another variable into the two changable variables, x and F; or balance the equation with another force (say, a mass on a spring so that F = mg).
You could also ask to determine the velocity and KE of the spring at any time or displacement of x. Or you could find the general solution to the differential equation of a harmonic oscillator, which is what you've got with a mass on a spring, and find sinusoidal motion in space, decaying exponentially with the damping constant. So it depends on what depth you need.
 
Hook's law is this: F= kx where F is the force applied to stretch or compress the spring, x is the distance the spring is stretched or compressed and k is the "spring constant". It basically says that the response of a spring is proportional to the force.

Your text may have F= -kx. The difference here is that F now is the force exerted BY the spring rather than the force exerted ON the spring ("equal and opposite").

Here are several "Hook's law" problems.

A spring with spring constant .4 cm/dyne has a force of 40 dynes applied to it (stretching it). How much does the spring stretch?

A force of 600 Newtons will compress a spring 0.5 meters. What is the spring constant of the spring?

A spring has spring constant 0.1 m/Newton. What force is necessary to stretch the spring by 2 meters?

A force of 40 Newtons will stretch a spring 0.1 meter. How far will a force of 80 Newtons stretch it?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 14 ·
Replies
14
Views
971
Replies
3
Views
1K
  • · Replies 35 ·
2
Replies
35
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 11 ·
Replies
11
Views
810
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K