Hooke's Law/Work done on a spring system

In summary, the conversation discusses a physics-based question with five parts, involving a block on a frictionless surface attached to a spring. The first problem must be solved correctly in order to answer the other four. The equations needed to solve the problem include Hooke's law and work/kinetic energy theorems. The person attempting the problem made a mistake by not including mass and acceleration in their equation. They are advised to set the work done by the force equal to the energy stored in the spring and integrate the differential equation to find the correct answer.
  • #1
ZeGinger
1
0

Homework Statement



I've been trying to answer a Physics-based question, and I can't seem to correctly answer the problem correctly. it is a five-part problem, and since I cannot answer the first problem correctly, the other four cannot be answered.
It is...

The block in the figure below lies on a horizontal frictionless surface and is attached to the free end of the spring, with a spring constant of 35 N/m. Initially, the spring is at its relaxed length and the block is stationary at position x = 0. Then an applied force with a constant magnitude of 3.2 N pulls the block in the positive direction of the x axis, stretching the spring until the block stops. Assume that the stopping point is reached.
(a) What is the position of the block?
(b) What is the work that has been done on the block by the applied force?
(c) What is the work that has been done on the block by the spring force?

I also need to find answers to these questions with regards to: 'during the block's displacement.'
(d) The block's position when its kinetic energy is maximum
(e) The value of that maximum kinetic energy.

Homework Equations



This particular problem utilizes Hooke's law and Work/Kinetic energy theorems. so far the only one I've used was Hooke's Law [itex] F = kx[/itex]
I'm aware that I will need to use work equations to solve for b and c W = .5kxi2 - .5kxf2

As for d and e, I've not a clue how to answer.

The Attempt at a Solution



When solving for part a, I simply manipulated Hooke's Law to solve using the aforementioned values (3.2N and 35N/M). I ended up getting 0.0914 meters, but it is considered incorrect.
Am I missing something in the equation?
 
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  • #2
You left out the mass times acceleration in your equation. Even though the mass has stopped, its acceleration at that point is backwards towards its initial position. You need to set the work done by the force equal to the energy stored in the spring, since the kinetic energy at maximum extension is zero. You will find that you underestimated by final extension by a factor of 2. Basically, the mass overshoots the equilibrium location you calculated.
 
  • #3
Hm, I'm not sure if I'm following. how would integrating Mass and Acceleration into the equation change the output of the problem? wouldn't the force value be all that's required...
or am I using mass and acceleration to calculated the fallback distance from the spring force?
 
  • #4
ZeGinger said:
Hm, I'm not sure if I'm following. how would integrating Mass and Acceleration into the equation change the output of the problem? wouldn't the force value be all that's required...
or am I using mass and acceleration to calculated the fallback distance from the spring force?
The differential equation is:
[tex]m\frac{d^2x}{dt^2}=F-kx[/tex]
Notice that the left hand side is not equal to zero (as you assumed). Solve the equation as a function of t, subject to the initial conditions x = 0 and dx/dt = 0 at time t = 0. Find the distance at which the velocity is again equal to zero. What do you get?
 
  • #5




Hello,

It seems like you are on the right track with your attempt at solving this problem. Hooke's Law, F = kx, is indeed the correct equation to use in this situation. However, it is important to remember that the force applied by the spring is equal and opposite to the applied force. In other words, the force applied by the spring will also be 3.2 N in the negative direction. This means that the total force acting on the block is 0 N, causing it to stop at some point along the x-axis.

To solve for the position of the block, you can use the equation F = kx. Since the total force is 0 N, you can set up the following equation: 0 = kx. Solving for x, you will get x = 0. This means that the block will stop at its initial position, x = 0.

For part (b), you can use the work equation W = Fd, where F is the applied force and d is the distance the block has been displaced. In this case, the distance would be the same as the position, x = 0. So the work done by the applied force would be 0 J.

For part (c), you can use the work equation W = 1/2kx^2. Plugging in the values, you will get W = 1/2(35)(0.0914)^2 = 0.143 J. This is the work done by the spring force on the block.

For parts (d) and (e), you can use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. Since the block starts and stops at the same position, its kinetic energy does not change. Therefore, the maximum kinetic energy will also be 0 J.

I hope this helps clarify the problem and how to approach it. Keep in mind that in physics, it is important to consider all the forces acting on an object and their directions. Good luck with the rest of your problem!
 

What is Hooke's Law?

Hooke's Law is a principle in physics that states the force needed to extend or compress a spring is directly proportional to the distance the spring is stretched or compressed.

What is the formula for Hooke's Law?

The formula for Hooke's Law is F = -kx, where F is the force applied, k is the spring constant, and x is the displacement of the spring from its equilibrium position.

What are the units for the spring constant?

The units for the spring constant depend on the units used for force and displacement. If the force is measured in Newtons (N) and displacement is measured in meters (m), then the units for the spring constant are N/m.

How is work done on a spring system calculated?

The work done on a spring system is calculated by multiplying the force applied by the displacement of the spring. This can be represented by the formula W = 1/2kx^2, where k is the spring constant and x is the displacement of the spring.

What are some real-life applications of Hooke's Law?

Hooke's Law is applicable in many real-life scenarios, such as in the design of springs for car suspensions, trampolines, and pogo sticks. It is also used in measuring the elasticity of materials and in the construction of various mechanical systems.

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