Finding the maximum compression of the spring

In summary, the conversation discusses a discrepancy in the solution for a problem due to a missing component in the question. The lecturer forgot to include the height of the object, leading to a different numerical answer. The suggested course of action is to show the solution to the lecturer and clarify the issue.
  • #1
Sean Harold
3
2
Homework Statement
11. Figure 7 shows a 3 kg box starts to move down an inclined plane with initial speed of 5 ms. It travels along the smooth surface and finally compresses the spring which has the spring constant of 400 Nm. Determine the maximum compression of the spring. [Given gravitational acceleration, g = 9.81 ms ²]

Correction:
Theta is not given and the height of the box from the ground is 0.8m
Relevant Equations
By using the conservation of energy,
The initial of total energy=The final of total energy
∑Eᵢ=∑Eғ
Kᵢ+Uᵢ=Kғ+Uғ
mghᵢ+1/2mvᵢ²=mghғ+1/2kx²
(3)(9.81)(0.8)+1/2(3)(5)²=(3)(9.81)(0)+1/2(400)(x)²
61.044=200x²
61.044/200=x²
x²=0.30522
x=0.5525m
Here is the answer I got for the question.
16319724068101635167086205417130.jpg
 
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  • #2
Where did the correction come from? It is not part of the statement of the problem that you posted. Your solution would be correct if the overall vertical drop of the mass were 0.8 m. However, the answer you got does not match the given answer so something is amiss.
 
  • #3
kuruman said:
Where did the correction come from? It is not part of the statement of the problem that you posted. Your solution would be correct if the overall vertical drop of the mass were 0.8 m. However, the answer you got does not match the given answer so something is amiss.
My lecturer told me that she had forgotten to put the height of the object in the question.
 
  • #4
Sean Harold said:
My lecturer told me that she had forgotten to put the height of the object in the question.
In that case, I think you should show your solution to your lecturer and ask her why your numerical answer does not match the given answer. Your method is correct and that's what counts.
 
  • Like
Likes Sean Harold
  • #5
kuruman said:
In that case, I think you should show your solution to your lecturer and ask her why your numerical answer does not match the given answer. Your method is correct and that's what counts.
Alright thanks for the help.
 

FAQ: Finding the maximum compression of the spring

What is the maximum compression of a spring?

The maximum compression of a spring refers to the point at which the spring can be compressed no further without causing permanent damage or deformation. It is the maximum amount of force that can be applied to the spring before it reaches its elastic limit.

How is the maximum compression of a spring determined?

The maximum compression of a spring can be determined by conducting experiments and measuring the amount of force needed to compress the spring until it can no longer return to its original shape. This can also be calculated using Hooke's Law, which states that the force applied to a spring is directly proportional to the amount of compression.

What factors can affect the maximum compression of a spring?

The maximum compression of a spring can be affected by various factors such as the material of the spring, its size and shape, and the amount of force applied. The temperature and environment in which the spring is used can also impact its maximum compression.

Why is it important to know the maximum compression of a spring?

Knowing the maximum compression of a spring is important in order to prevent damage or failure of the spring. It can also help in determining the appropriate amount of force to apply in different situations, such as in mechanical systems or experiments.

Can the maximum compression of a spring be increased?

The maximum compression of a spring is determined by its physical properties and cannot be increased beyond its elastic limit. However, the maximum compression can be increased by using multiple springs in a series or by adjusting the design and material of the spring to increase its elasticity.

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