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Q.

Consider a block kept on an inclined plane (inclined at 45°) as shown in the figure. If the force required to just push it up the incline is 2 times the force required to just prevent it from sliding down, the coefficient of friction between the block and inclined plane $(μ)$ is equal to

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a

0.50

b

0.25

c

0.60

d

0.33

answer is B.

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Detailed Solution

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Let F₁ be the force required to just prevent it from sliding $F_{1} = mgsin θ - μmgcos θ$
Let F₂ be the minimum force to push it up the incline
$\begin{array}{l} F_{2} = mgsin θ + μmgcos θ \\ F_{2} = 2 F_{1} \\ mgsin θ + μmgcos θ = 2 (mgsin θ - μmgcos θ) \\ 1 + μ = 2 (1 - μ) \\ μ = \frac{1}{3} \end{array}$

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