The upper $\left(\frac{1}{n}\right)^{\text {th }}$ of an inclined plane is smooth and the remaining lower…

The upper $\left(\frac{1}{n}\right)^{\text {th }}$ of an inclined plane is smooth and the remaining lower part is rough with coefficient of friction $\mu_k \cdot$ If a body starting from rest at the top of the inclined plane will again come to rest at the bottom of the plane, then the angle of inclination of the inclined plane is
  1. $\sin ^{-1}\left[\left(\frac{n}{n-1}\right) \mu_k\right]$
  2. $\sin ^{-1}\left[\left(\frac{n-1}{n}\right) \mu_k\right]$
  3. $\tan ^{-1}\left[\left(\frac{n}{n-1}\right) \mu_k\right]$
  4. $\tan ^{-1}\left[\left(\frac{n-1}{n}\right) \mu_k\right]$

Solution


$\begin{aligned} & \mathrm{W}_{\mathrm{g}}=\mathrm{mgh}=\mathrm{mgl} \sin \theta \\ & \mathrm{~W}_{\mathrm{f}}=-\left(\mu_{\mathrm{k}} \mathrm{mg} \cos \theta\right) \cdot 1\left(1-\frac{1}{\mathrm{n}}\right) \\ & \mathrm{W}_{\mathrm{N}}=0 \end{aligned}$
By work energy theorem, $\begin{aligned} & \mathrm{W}_{\mathrm{g}}+\mathrm{W}_{\mathrm{f}}+\mathrm{W}_{\mathrm{N}}=\Delta \mathrm{k}=0 \\ & \Rightarrow \mathrm{mg} l \sin \theta-\left(\mu_{\mathrm{k}} \mathrm{mg} \cos \theta\right) \cdot l\left(\frac{\mathrm{n}-1}{\mathrm{n}}\right)+0=0 \\ & \therefore \quad \theta=\tan ^{-1}\left[\left(\frac{\mathrm{n}-1}{\mathrm{n}}\right) \cdot \mu_{\mathrm{k}}\right] \end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 1)

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