The upper half of an inclined plane with incl | vedclass

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Question

$K.E$ of a block at point $P$ is given as

$\mathrm{K.E}=m g-\frac{l}{2} \sin \phi$

At point $C,$ $K.E$ of the block will be zero, therefore $K .

Vedclass · Accepted Answer

$2 	an \phi$

Vedclass · Answer

$K.E$ of a block at point $P$ is given as

$\mathrm{K.E}=m g-\frac{l}{2} \sin \phi$

At point $C,$ $K.E$ of the block will be zero, therefore $K . E$ is lost in work done by friction.

$W=\mu N \frac{l}{2}=\mu m g \frac{l}{2} \cos \phi$

By energy conservation at point $B$ and $C;$

At B total energy $=K . E+P . E=K . E+m g h=K . E+m g\left(\frac{l}{2} \sin \phiight)=m g\left(\frac{l}{2} \sin \phiight)$

$+m g\left(\frac{l}{2} \sin \phiight)=m g l \sin \phi$

Now, $W=$ total energy at $B$

since potential energy at $C$ is zero.

$\mu\left(\frac{1}{2}ight) m g l \cos \phi=m g l \sin \phi$

$\mu=\frac{2 \sin \phi}{\cos \phi}=2 	an \phi$

The upper half of an inclined plane with inclination $\phi$ is perfectly smooth, while the lower half is rough. $A$ body starting from rest at the top will again come to rest at the bottom, if the coefficient of friction for the lower half is given by-

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