AP EAMCET · PHYSICS · Work Power Energy
The work done in stretching a spring of natural length 25 \(\mathrm{cm}\) and spring constant \(50 \mathrm{Nm}^{-1}\) from \(50 \mathrm{~cm}\) to \(60 \mathrm{~cm}\) is
- A \(1.5 \mathrm{~J}\)
- B \(2 \mathrm{~J}\)
- C \(3.5 \mathrm{~J}\)
- D \(5 \mathrm{~J}\)
Answer & Solution
Correct Answer
(A) \(1.5 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
\(\Delta \mathrm{x}_{\mathrm{i}}=(50-25) \mathrm{cm}=25 \mathrm{~cm}\) \(\Delta \mathrm{x}_{\mathrm{f}}=(60-25) \mathrm{cm}=35 \mathrm{~cm}\) So, work done \(=\Delta U\)…
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