TS EAMCET · Physics · Work Power Energy
A mechanical system consists two springs of stiffness coefficients \(k_1\) and \(k_2\) are connected in series. The minimum work to be performed on the system to stretch it by \(\Delta\) is
- A \(\frac{1}{2}\left(\frac{k_1 k_2}{k_1+k_2}\right) \Delta^2\)
- B \(k_1 k_2 \Delta^2\)
- C \(\left(\frac{k_1 k_2}{k_1+k_2}\right) \Delta^{\prime 2}\)
- D \(\left(\frac{k_1 k_2}{k_1+k_2}\right) \Delta\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2}\left(\frac{k_1 k_2}{k_1+k_2}\right) \Delta^2\)
Step-by-step Solution
Detailed explanation
\(K_{\text {eq }}(\) in series \()=\frac{k_2 k_2}{k_2+k_2}\) Work done to stretch springs with elongation \(\Delta l\) is \[ W=\frac{1}{2} K_{\text {eq }} \cdot \Delta t^2=\frac{1}{2}\left(\frac{k_2 k_2}{k_1+k_2}\right) \cdot \Delta l^2 \]
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