Starting from rest, time taken by a body to slide down a 45° inclined plane with friction, is twice the time taken to slide down the same distance in absence of friction.Find coefficient of friction?

Let's denote the time taken to slide down the inclined plane with friction as \(t_f\) and the time taken to slide down the same distance in absence of friction as \(t_{nf}\).

Given that \(t_f = 2t_{nf}\).

We know that the time taken for an object to slide down an inclined plane with friction is given by:

\[t_f = \sqrt{\frac{2h}{g\sin\theta(\mu + \cos\theta)}}\]

where:
\(h\) = height of the inclined plane,
\(g\) = acceleration due to gravity,
\(\theta\) = angle of inclination,
\(\mu\) = coefficient of friction.

The time taken for an object to slide down an inclined plane without friction is given by:

\[t_{nf} = \sqrt{\frac{2h}{g\sin\theta}}\]

Given that \(t_f = 2t_{nf}\), we can substitute these equations into the given equation:

\[\sqrt{\frac{2h}{g\sin\theta(\mu + \cos\theta)}} = 2\sqrt{\frac{2h}{g\sin\theta}}\]

Squaring both sides, we get:

\[\frac{2h}{g\sin\theta(\mu + \cos\theta)} = 4\frac{2h}{g\sin\theta}\]

Simplifying, we get:

\[\mu + \cos\theta = 4\]

Given that the angle of inclination is 45°, we can substitute \(\theta = 45°\) into the equation:

\[\mu + \cos(45°) = 4\]

\[\mu + \frac{1}{\sqrt{2}} = 4\]

\[\mu = 4 - \frac{1}{\sqrt{2}}\]

\[\mu = 4 - \frac{\sqrt{2}}{2}\]

\[\mu = \frac{8 - \sqrt{2}}{2}\]

Therefore, the coefficient of friction is \(\frac{8 - \sqrt{2}}{2}\).