First slide

Friction on inclined plane of angle more than angle of repose

Question

The upper half of an inclined plane of inclination $θ$ is perfectly smooth while the lower half 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 the lower half of the plane is given by

Moderate

A

$μ$ = 2 tan $θ$
B

$μ$ = tan $θ$
C

$μ$ = 2 / tan $θ$
D

$μ$ = 1 / tan $θ$

Solution

Suppose length of plane is L. When block descends the plane, rise in kinetic energy = fall in potential energy
= mg Lsin $θ$
Work done by friction
= $μ$ x (Reaction) x Distance
$= 0 + μ (m g \cos θ) \times (L / 2)$
$= μ (m g \cos θ) \times (L / 2)$
Now, work done = change in KE
$m g L \sin θ = μ (m g \cos θ) \times (L / 2) \Rightarrow \tan θ = μ / 2$
$∴$ $μ$ = 2 tan $θ$

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