Calculating Velocity using Hooke's Law: A Simple Question

In summary: I'll have to look into that.In summary, the block moves at a speed of 4.00 N when located at the equilibrium point of the spring.
  • #1
unnamedplayer
6
0
Hi again, all. I thought of another question that I need help with. Again I have the answer but I'm not sure how they got to it. The question is:

A 1.6kg block is supported on a horizontal surface. The block is attached to a sping with k = 1000 N/m. The block compresses the spring x = 0.02 m and is then released from rest. The surface provides 4.00 N constant frictional force. Determine the speed of the block when located at spring's equilibrium point.

Now, I know to use W = (1/2)mv^2 - (1/2)mv^2 to find the velocity but I'm having trouble understanding how they got the total work. What they did was find the work done by the spring on the block and then find the work done by the friction and subtract the two to get the net work done.

But I tried to do it by calculating the spring force (which I got 20N for), subtract the friction so net force is 16 N and then multiply that by the distance 0.02 to get the net work done. But I get the wrong answer and I don't understand why. Maybe I am not calculating the spring force right? I'm using F = -kx.

Any help as to what I'm doing wrong would be greatly appreciated!
 
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  • #2
maybe you need to take in consideration acceleration due to gravity?
the mass 1.6 would be acted upon by gravity when it is let go right? so there you have downwards force
 
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  • #3
unnamedplayer said:
But I tried to do it by calculating the spring force (which I got 20N for), subtract the friction so net force is 16 N and then multiply that by the distance 0.02 to get the net work done. But I get the wrong answer and I don't understand why. Maybe I am not calculating the spring force right? I'm using F = -kx.
Any help as to what I'm doing wrong would be greatly appreciated!
Force isn't constant. You'll need calculus to solve it that way.

Potential energy of a spring:
[tex]E = \frac{1}{2}kx^2[/tex]

alias25 said:
maybe you need to take in consideration acceleration due to gravity?
the mass 1.6 would be acted upon by gravity when it is let go right? so there you have downwards force
It's directed downwards. But then there's the support force of the surface upwards, which is equal but opposite to the force caused by gravity. Thus we only need to consider horizontal motion.
 
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  • #4
So if the force isn't constant I have to find the work done by each force then add or subtract them to get the net work? Otherwise if the force is constant I can use the method I tried using in which I just find the net force and then multiply by the distance to find the work right?
 
  • #5
As Päällikkö was trying to say (I think) you don't have to workout the individual forces, just the energies. In the initial position there is only potential energy from the spring, at the equilibrium there is only kinetic energy so you can write that:
[tex] \frac{1}{2}mv_{f}^2 = \frac{1}{2}kx_{i}^2[/tex]
 
  • #6
I don't think so Daniel, you're applying the Conservation of energy without taking care of friction.

[tex] \Delta E = W_{friction} [/tex]
 
  • #7
Sorry, I didn't notice the friction.
 

What is Hooke's Law?

Hooke's Law is a scientific principle that states the force needed to extend or compress a spring is directly proportional to the distance the spring is extended or compressed. In simpler terms, it explains the relationship between the force applied to an elastic object and the resulting deformation.

What is the equation for Hooke's Law?

The equation for Hooke's Law is F = -kx, where F is the applied force, k is the spring constant, and x is the displacement from the equilibrium position. This equation shows the direct proportionality between force and displacement.

How is Hooke's Law used in real life?

Hooke's Law has many practical applications, such as in the design of springs for various mechanical devices, like car suspensions or pogo sticks. It is also used in materials testing to determine the elasticity and strength of different materials, such as rubber or metals.

What is the spring constant?

The spring constant, represented by the letter k, is a measure of the stiffness of a spring. It is a constant value that relates the amount of force required to extend or compress a spring by a certain distance. A higher spring constant indicates a stiffer spring, while a lower spring constant indicates a more flexible spring.

What are the limitations of Hooke's Law?

Hooke's Law is based on certain assumptions, such as the material being tested follows a linear elastic behavior and the force applied is not enough to permanently deform the material. In reality, materials often exhibit non-linear behavior and can reach a point of permanent deformation. Additionally, Hooke's Law is only applicable to springs and certain elastic materials, and may not accurately represent the behavior of more complex materials or structures.

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