Question

A body takes n times, the time to slide down a rough inclined plane as it takes to slide down the same inclined plane when it is perfectly frictionless. The coefficient of kinetic friction between the body and the plane for an angle of inclination of 45° is given by μ

• A. 1- $\frac { 1 } { \mathrm { n } }$
• B. $\frac { 1 } { \mathrm { n } }$
• C. $\left( 1 - \frac { 1 } { n ^ { 2 } } \right)$
• D. $\left( \frac { 1 } { n ^ { 2 } } - 1 \right)$

Answer 1

Answer: (C)

$\left( 1 - \frac { 1 } { n ^ { 2 } } \right)$

Answer 2

Let a be the acceleration down the rough plane and a' be the acceleration down the frictionless plane. Taking L as the length of the inclined plane, we get

a = g(sinθ - μcos θ)

= $g \left( \frac { 1 } { \sqrt { 2 } } - \frac { \mu } { \sqrt { 2 } } \right)$ ( θ = 45⁰)

and a' = g sin θ = g $\frac { 1 } { \sqrt { 2 } }$

Then L = $\frac { 1 } { 2 }$ at₁² = $\frac { 1 } { 2 }$ a't₂²

or $\frac { 1 } { 2 } \mathrm {~g} \left( \frac { 1 } { \sqrt { 2 } } - \frac { \mu } { \sqrt { 2 } } \right) \mathrm { t } _ { 1 } ^ { 2 } = \frac { 1 } { 2 } \frac { \mathrm { g } } { \sqrt { 2 } } \mathrm { t } _ { 2 } ^ { 2 }$

But t₁ = nt₂ (given)

∴ $\frac { 1 } { 2 } \mathrm {~g} \left( \frac { 1 } { \sqrt { 2 } } - \frac { \mu } { \sqrt { 2 } } \right) \mathrm { n } ^ { 2 } \mathrm { t } _ { 2 } ^ { 2 } = \frac { 1 } { 2 } \frac { \mathrm { g } } { \sqrt { 2 } } \mathrm { t } _ { 2 } ^ { 2 }$ or 1 = (1 - μ)n² or μ = $\left( 1 - \frac { 1 } { n ^ { 2 } } \right)$