Question

A block of mass m slides down an inclined right angled trough. If the coefficient of kinetic friction between the block and the trough is μ_k, acceleration of the block down the plane is

• A. g(sin θ - 2μ_kcosθ)
• B. g (sin θ + 2μ_kcosθ)
• C. g(sin θ+ √2μ_kcosθ)
• D. g(sin θ- μ_kcosθ)

Answer 1

Answer: (C)

g(sin θ+ √2μ_kcosθ)

Answer 2

Note

(i) m acts downward

(ii) frictional forces up the plane

(iii) reactions N as shown

Then mg sin θ - 2μ_kN = ma

( there are 2 surfaces)

also mg cos θ - $\sqrt { 2 }$ N = 0

From (i) and (ii)

mg sin θ - 2μ_k $\frac { m g \cos \theta } { \sqrt { 2 } }$ = ma

or a = g(sin θ - $\sqrt { 2 }$ μ_k cos θ)