A block of mass 5.0 kg slides down from the top of an inclined plane of length 3

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Question -

A block of mass 5.0 kg slides down from the top of an inclined plane of length 3 m. The first 1 m of the plane is smooth and the next 2 m is rough. The block is released from rest and again comes to rest at the bottom of the plane. If the plane is inclined at $30^{\circ}$ with the horizontal, find the coefficient of friction on the rough portion.

Options:

A .

\frac{2}{\sqrt{3}}

B .

\frac{\sqrt{3}}{2}

C .

\frac{\sqrt{3}}{4}

D .

\frac{\sqrt{3}}{5}

Answer: Option B
:
B

Force Method:

Form P to Q: $v^{2} = O^{2} + 2 a_{1} S_{1}$

Where $a_{1} = g s i n 30^{\circ}, s_{1} = 1 m$

And form Q to R: $O^{2} = v^{2} + 2 a_{2} s_{2}$

Where $a_{2} = g s i n 30^{\circ} - μ g c o s 30^{\circ}, s_{2} = 2 m$

Solve to get $μ = \frac{\sqrt{3}}{2}$

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