JEE Mains · Physics · STD 11 - 4.2 friction
A block of mass \(2\,kg\) moving on a horizontal surface with speed of \(4\,ms ^{-1}\) enters a rough surface ranging from \(x =0.5\,m\) to \(x =1.5\,m\). The retarding force in this range of rough surface is related to distance by \(F =- kx\) where \(k =12\,Nm ^{-1}\). The speed of the block as it just crosses the rough surface will be ........... \(\,ms ^{-1}\)
- A \(0\)
- B \(1.5\)
- C \(2.0\)
- D \(2.5\)
Answer & Solution
Correct Answer
(C) \(2.0\)
Step-by-step Solution
Detailed explanation
\(a=\frac{- kx }{2}=\frac{-12 x }{2}=-6 x\) \(\frac{ v d v }{ dx }=-6 x\) \(\int \limits_{4}^{*} vdv =-\int \limits_{\frac{1}{2}}^{ s / 2} 6 xdx\) \(\frac{ v ^{2}-4^{2}}{2}=-\frac{6}{2}\left[\left(\frac{3}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}\right]\)…
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