Question

The upper half of an inclined plane of inclination (theta) is perfectly smooth with the lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and lower half of the plane is given by​​

Answer 1

Answer:

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The upper half of an inclined plane

of inclination (theta)

is perfectly smooth with the lower half is rough

and a block starting from rest at the top of the plane will again come to rest at the bottom

to happen this

acceleration of the block on the smooth surface must be equal to the retardation of block on the rough surface

we know that,

acceleration on a smooth incline = gsin@

where,

g = gravitational acceleration

@ = inclination of the plane

and retardation on a rough inclined

= g(sin@ + ú cos@)

so,

acceleration = - g(sin@ + ú cos@)

where,

ú = coefficient of kinetic friction

so,

g sin@ = -g(sin@ + ú cos@)

sin@ = -sin@ + ú cos@

sin@ + sin@ = ú cos@

ú = 2 sin@/cos@

ú = 2tan@

so,

Coefficient of kinetic friction

= 2 tan@

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Explanation:

Answer 2

Answer:

μ=2tanθ

Explanation:

The acceleration of the block while it is sliding down the upper half of the inclined plane is gsinθ.

If μ is the coefficient of kinetic friction between the block and the lower half of the plane,the retardation of the block while it is sliding down the lower half = −(gsinθ−μcosθ)

For the block to come to rest at the bottom of the inclined plane,the acceleration in the first half must be equal to the retardation in the second half,i.e.

gsinθ=−(gsinθ−μgcosθ)

μcosθ=2sinθ

μ=2tanθ

Hence the correct answer is μ=2tanθ