Conservation of energy spring problem

In summary, a block of mass 2.00 kg attached to a spring of force constant 500 N/M is pulled to a position of 0.05m and released from rest. The speed of the block as it passes equilibrium can be found by using the equation 1/2mv^2 = 1/2k(x_o)^2, where x_o is the initial amount the spring is stretched. This is because the total energy of the system remains constant when there is no friction.
  • #1
whynot314
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Homework Statement


A block of mass m=2.00 kg is attached to a spring of force constant κ=500 N/M. the block is pulled to a position xi=.05m and released from rest. Find the speed the block has as it has passed equilibrium. Assume frictionless.

Homework Equations


ΔEsystem=ƩT(transfer)
Δk+ΔU=W

The Attempt at a Solution


I am assuming that this is a non isolated system.So I am having trouble setting up this equation I know work done by spring=1/2kX^2 and I know what Δk is, but Do I need to include the difference in elastic potential? I have seen this worked out without the elastic potential. why is it not included? and they just solved 1/2mv^2=1/2kx^2?
 
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  • #2
You know that when the mass vibrates and there is no friction the total energy of the system remains constant. The total energy E is,

E = 1/2mv^2 + 1/2kx^2 = constant = 1/2k(x_o)^2 where x_o is the initial amount the spring is stretched.

1/2mv^2 + 1/2kx^2 = 1/2k(x_o)^2 but we want to know v when x = 0 so,

1/2mv^2 = 1/2k(x_o)^2

So use the last equation you wrote but remember x is the initial amount the spring was stretched.
 

1. How does the conservation of energy apply to a spring problem?

The conservation of energy states that energy cannot be created or destroyed, but can only be converted from one form to another. In the case of a spring problem, the potential energy stored in the spring is converted into kinetic energy as the spring is compressed or stretched. This energy is then conserved as it is released and converted back to potential energy.

2. What factors affect the conservation of energy in a spring problem?

The conservation of energy in a spring problem is affected by the mass of the object attached to the spring, the distance the spring is compressed or stretched, and the spring constant which determines the stiffness of the spring. These factors all contribute to the total potential and kinetic energy in the system.

3. Is the conservation of energy always applicable to a spring problem?

Yes, the conservation of energy is always applicable to a spring problem as long as there are no external forces acting on the system. This means that the system is isolated and there is no friction or air resistance that could cause a loss of energy.

4. Can the conservation of energy be used to solve a spring problem?

Yes, the conservation of energy can be used to solve a spring problem. By setting the initial potential energy equal to the final kinetic energy, the equation for conservation of energy can be used to find the displacement or velocity of the object attached to the spring.

5. Are there any real-world applications of the conservation of energy in spring problems?

Yes, there are many real-world applications of the conservation of energy in spring problems. For example, it is used in the design of shock absorbers in cars, in the operation of certain types of clocks, and in the functioning of trampolines. Understanding the conservation of energy in spring problems is crucial in many fields such as engineering, physics, and mechanics.

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