Conservation of Energy- spring

In summary, the conversation discusses solving a physics problem involving a 6kg block sliding on a frictionless surface and hitting a bumper that compresses a spring. The student uses the Law of Conservation of Energy and assumes there is no Gravitational Potential Energy, leading to an incorrect answer. Upon receiving help, the student realizes their mistake in the units of the spring potential energy term.
  • #1
Maiia
79
0

Homework Statement


A 6kg block slides on a horizontal frictionless surface with a speed of 1.2 m/s. It is brought momentarily to rest when it hits a bumper that compresses a spring.

Heres a picture of the problem.
springproblem.jpg


When I was doing this problem, I thought there were only two different energies: the Kinetic Energy and the Elastic Potential Energy. I assumed there was no Gravitational PE b/c there was no height involved. Therefore, I assumed (according to Law of Conservation of Energy) that you could set the Kinetic Energy equal to EPE.
(1/2)mv^2 + (1/2)kx^2= (1/2)mv2^2 + (1/2)kx2^2 which reduces to
(1/2)mv^2= (1/2) k(-Y2)
I am using -Y to denote the compression of the spring
if I plug in and solve
(1/2)(6kg)(1.2m/s ^2)= (1/2) (4)(Y2)
I get 1.341640786 m as the amount of compression. However, this answer is wrong. Can someone tell me what I did wrong?
 
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  • #2
Maiia said:

Homework Statement


A 6kg block slides on a horizontal frictionless surface with a speed of 1.2 m/s. It is brought momentarily to rest when it hits a bumper that compresses a spring. When I was doing this problem, I thought there were only two different energies: the Kinetic Energy and the Elastic Potential Energy. I assumed there was no Gravitational PE b/c there was no height involved. Therefore, I assumed (according to Law of Conservation of Energy) that you could set the Kinetic Energy equal to EPE.
(1/2)mv^2 + (1/2)kx^2= (1/2)mv2^2 + (1/2)kx2^2 which reduces to
(1/2)mv^2= (1/2) k(-Y2)
I am using -Y to denote the compression of the spring
if I plug in and solve
(1/2)(6kg)(1.2m/s ^2)= (1/2) (4)(Y2)
I get 1.341640786 m as the amount of compression. However, this answer is wrong. Can someone tell me what I did wrong?

Check your units. Spring constant is 4K*N/m
 
  • #3
Oh wow that was stupid of me. Thanks!
 
  • #4
Your understanding of the physics is correct, but perhaps you should check if your spring potential energy term with Y_2 is correct. Energy has units of Newton x meter, and the spring constant has units of Newton per meter.
 

1. What is the law of conservation of energy?

The law of conservation of energy states that energy cannot be created or destroyed, but it can only be transferred or transformed from one form to another.

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

A spring possesses potential energy when it is compressed or stretched. As it is released, this potential energy is converted into kinetic energy, and the total energy of the system remains constant.

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

The conservation of energy in a spring is influenced by its mass, spring constant, and amplitude of oscillation. A heavier mass, higher spring constant, and larger amplitude will result in a greater amount of potential and kinetic energy in the system.

4. Can the conservation of energy be violated in a spring system?

No, the conservation of energy is a fundamental law of physics and cannot be violated. In a spring system, energy may be lost due to external factors such as friction, but the total amount of energy in the system will remain constant.

5. How is the conservation of energy demonstrated in a real-life application of a spring?

A common real-life application of the conservation of energy in a spring can be seen in a pogo stick. As the user jumps up and down on the pogo stick, the potential energy of the compressed spring is converted into kinetic energy, allowing the user to continue bouncing without expending additional energy.

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