Question

A block of mass m is kept on inclined plane of a lift moving down with acceleration of what should be coefficient of friction to let the block move down the inclined plane with constant velocity?

Answer 1

From the free body diagrams as in the diagram,

mg cos θ − ma cos θ=N

⇒N=m(g−a) cos θ

∴Ffri=μN=μ{m(g−a) cos θ}

⇒mg sin θ − ma sin θ

=μ{m(g−a) cos θ}

⇒sin θ=μcos θ

⇒μ=sin θ/cos θ=tan θ

⇒μ=tan θ

Answer 2

Explanation:

so this is the correct answer