The ratio of magnitude of work done in moving slowly a block of mass m on a rough inclined surface of inclination - - 45-tan- from bottom to top and from top to bottom is 5- The coefficient of friction between the block and surface is-

The correct option is B 2/3Apply work energy theorem #160; W_ B T + W _ mg + W _ f =0 W_ B T mgsinθ× 5 μ mgcosθ× 5=0 W_ B T =( mgsinθ +μ mgc ...

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Standard XII

Physics

Centripetal Acceleration

The ratio of ...

Question

The ratio of magnitude of work done in moving slowly a block of mass m on a rough inclined surface of inclination $θ$ = $45^{\circ} (t a n θ > μ)$ from bottom to top and from top to bottom is 5. The coefficient of friction between the block and surface is:

1/2

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2/3

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2/5

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4/5

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Solution

The correct option is B 2/3
Apply work energy theorem
$W_{B - T} + W_{m g} + W_{f} = 0 W_{B - T} - m g s i n θ \times 5 - μ m g c o s θ \times 5 = 0 W_{B - T} = (m g s i n θ + μ m g c o s θ) 5 S i m i l a r l y W_{B - T} + m g s i n θ \times 5 - μ m g c o s θ \times 5 = 0 W_{B - T} = (m g s i n θ - μ m g c o s θ) 5 \frac{(m g s i n θ + μ m g c o s θ)}{- (m g s i n θ + μ m g c o s θ)} = 5 4 m g s i n θ = 6 μ m g c o s θ μ = \frac{2}{3}$

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The ratio of magnitude of work done in moving slowly a block of mass m on a rough inclined surface of inclination θ = 45∘(tanθ>μ) from bottom to top and from top to bottom is 5. The coefficient of friction between the block and surface is:

The correct option is B 2/3Apply work energy theorem WB−T+Wmg+Wf=0WB−T−mgsinθ×5−μmgcosθ×5=0WB−T=(mgsinθ+μmgcosθ)5SimilarlyWB−T+mgsinθ×5−μmgcosθ×5=0WB−T=(mgsinθ−μmgcosθ)5(mgsinθ+μmgcosθ)−(mgsinθ+μmgcosθ)=54mgsinθ=6μmgcosθμ=23

The ratio of magnitude of work done in moving slowly a block of mass m on a rough inclined surface of inclination $θ$ = $45^{\circ} (t a n θ > μ)$ from bottom to top and from top to bottom is 5. The coefficient of friction between the block and surface is: