JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
A body of mass \(m= 10^{-2}\) \(kg\) is moving in a medium and experiences a frictional force \(F= -kv^2\). Its initial speed is \(v_0= 10\) \(ms^{-1}\). If, after \(10\; s\), its energy is \(\frac{1}{8}\) \(mv_0^2\) the value of \(k\) will be
- A \(10^{-3}\) \(kg m^{-1}\)
- B \(10^{-3}\) \(kg s^{-1}\)
- C \(10^{-4}\) \( kg m^{-1}\)
- D \(10^{-1}\) \(kg m^{-1} s^{-1}\)
Answer & Solution
Correct Answer
(C) \(10^{-4}\) \( kg m^{-1}\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{2} m v_{f}^{2}=\frac{1}{8} m v_{0}^{2} \Rightarrow v_{f}=\frac{v_{0}}{2}\) Now, \(\frac{\mathrm{mdv}}{\mathrm{dt}}=-\mathrm{kv}^{2}\)…
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