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If a close coiled helical spring absorbs 30 N-mm of energy while extending by 5mm, its stiffness will be ______.
- a)2 N/mm
- b)4 N/mm
- c)2.4 N/mm
- d)10 N/mm
Correct answer is option 'C'. Can you explain this answer?
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Concept:
The stiffness of a spring is a measure of its ability to resist deformation when a force is applied to it. It is defined as the ratio of the force applied to the spring to the resulting displacement.
Given:
Energy absorbed by the spring = 30 N-mm
Extension of the spring = 5 mm
Calculation:
The energy absorbed by a spring can be calculated using the formula:
Energy = (1/2) * k * x^2
Where:
- Energy is the energy absorbed by the spring (in N-mm)
- k is the stiffness of the spring (in N/mm)
- x is the extension of the spring (in mm)
- Energy is the energy absorbed by the spring (in N-mm)
- k is the stiffness of the spring (in N/mm)
- x is the extension of the spring (in mm)
Substituting the given values into the formula, we have:
30 N-mm = (1/2) * k * (5 mm)^2
Simplifying the equation:
30 N-mm = (1/2) * k * 25 mm^2
Dividing both sides of the equation by (1/2) * 25 mm^2:
30 N-mm / ((1/2) * 25 mm^2) = k
Simplifying further:
30 N-mm / (12.5 mm^2) = k
Expressing the stiffness in N/mm:
30 N-mm / (12.5 mm^2) = k
k = 2.4 N/mm
Answer:
The stiffness of the close coiled helical spring is 2.4 N/mm.
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If a close coiled helical spring absorbs 30 N-mm of energy while extending by 5mm, its stiffness will be ______.a)2 N/mmb)4 N/mmc)2.4 N/mmd)10 N/mmCorrect answer is option 'C'. Can you explain this answer?
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