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To compress a helical spring that is already compressed, where
Lo = 5.0 cm = 0.05 m
Li = 4.5 cm = 0.045 m
Lf = 3.5 cm = 0.035 m
k = 18,000 N/m
The work done will be:
- a)+1.8 Nm
- b)-1.6 Nm
- c)-1.8 Nm
- d)+1.6 Nm
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
To compress a helical spring that is already compressed, whereLo = 5....
Step 1: using eq.: x1 = Li – Lo, x2 = Lf – Lo; Calculate the displacements (x1) and (x2) as
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x1 = Li − Lo = (0.045 m) − (0.05 m)
= −0.005 m
x2 = Lf − Lo = (0.035 m) − (0.05 m)
= −0.015 m
Step 2: Substitute the displacements (x1) and (x2) found in step 1 in Equation to give the work done as
Work1→2 = (1/2) k(x21 − x22)
= 1/2(18,000 N/m) × ((−0.005 m) 2 − (−0.015 m) 2)
= 1/2(18,000 N/m) × ((0.000025 − 0.000225) m2)
= 1/2(18,000 N/m)x (−0.0002 m2)
= −1.8 N · m = −180 N · cm
The negative sign on the work done means work was done on the spring.
Most Upvoted Answer
To compress a helical spring that is already compressed, whereLo = 5....
The Problem:
A helical spring is already compressed with an initial length (Lo) of 0.05 m and a final length (Li) of 0.045 m. The spring constant (k) is given as 18,000 N/m. We need to determine the work done on the spring during this compression.
Approach:
To find the work done on the spring, we can use the formula:
W = (1/2)k(Lf^2 - Li^2)
Solution:
Given:
Initial length (Lo) = 0.05 m
Final length (Li) = 0.045 m
Spring constant (k) = 18,000 N/m
To calculate the work done, we need to find the change in length of the spring (ΔL) and substitute it into the formula.
Change in length (ΔL) = Lo - Li
= 0.05 m - 0.045 m
= 0.005 m
Now, we can substitute the values into the formula for work done:
W = (1/2)k(ΔL^2)
= (1/2)(18,000 N/m)(0.005 m)^2
= (1/2)(18,000 N/m)(0.000025 m^2)
= 0.225 Nm
Therefore, the work done on the spring during compression is 0.225 Nm.
Answer:
The correct option is (c) -1.8 Nm.
A helical spring is already compressed with an initial length (Lo) of 0.05 m and a final length (Li) of 0.045 m. The spring constant (k) is given as 18,000 N/m. We need to determine the work done on the spring during this compression.
Approach:
To find the work done on the spring, we can use the formula:
W = (1/2)k(Lf^2 - Li^2)
Solution:
Given:
Initial length (Lo) = 0.05 m
Final length (Li) = 0.045 m
Spring constant (k) = 18,000 N/m
To calculate the work done, we need to find the change in length of the spring (ΔL) and substitute it into the formula.
Change in length (ΔL) = Lo - Li
= 0.05 m - 0.045 m
= 0.005 m
Now, we can substitute the values into the formula for work done:
W = (1/2)k(ΔL^2)
= (1/2)(18,000 N/m)(0.005 m)^2
= (1/2)(18,000 N/m)(0.000025 m^2)
= 0.225 Nm
Therefore, the work done on the spring during compression is 0.225 Nm.
Answer:
The correct option is (c) -1.8 Nm.
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To compress a helical spring that is already compressed, whereLo = 5.0 cm = 0.05 mLi = 4.5 cm = 0.045 mLf = 3.5 cm = 0.035 mk = 18,000 N/mThe work done will be:a)+1.8 Nmb)-1.6 Nmc)-1.8 Nmd)+1.6 NmCorrect answer is option 'C'. Can you explain this answer?
Question Description
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To compress a helical spring that is already compressed, whereLo = 5.0 cm = 0.05 mLi = 4.5 cm = 0.045 mLf = 3.5 cm = 0.035 mk = 18,000 N/mThe work done will be:a)+1.8 Nmb)-1.6 Nmc)-1.8 Nmd)+1.6 NmCorrect answer is option 'C'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about To compress a helical spring that is already compressed, whereLo = 5.0 cm = 0.05 mLi = 4.5 cm = 0.045 mLf = 3.5 cm = 0.035 mk = 18,000 N/mThe work done will be:a)+1.8 Nmb)-1.6 Nmc)-1.8 Nmd)+1.6 NmCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for To compress a helical spring that is already compressed, whereLo = 5.0 cm = 0.05 mLi = 4.5 cm = 0.045 mLf = 3.5 cm = 0.035 mk = 18,000 N/mThe work done will be:a)+1.8 Nmb)-1.6 Nmc)-1.8 Nmd)+1.6 NmCorrect answer is option 'C'. Can you explain this answer?.
Solutions for To compress a helical spring that is already compressed, whereLo = 5.0 cm = 0.05 mLi = 4.5 cm = 0.045 mLf = 3.5 cm = 0.035 mk = 18,000 N/mThe work done will be:a)+1.8 Nmb)-1.6 Nmc)-1.8 Nmd)+1.6 NmCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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