A body slides from the top of an inclined plane of height h…

A body slides from the top of an inclined plane of height $h = 6\ \mathrm{m}$ and, after continuing along a horizontal path, stops at a point 25 m away horizontally from its starting point. The coefficient of friction is the same on the incline and on the horizontal path. Find it.

Explanation of Method

Friction coefficient from the stopping distance.

Step 1: Friction work depends only on horizontal distance

On the incline of angle $\theta$ and length $s_1$, friction does work $\mu mg\cos\theta \cdot s_1 = \mu mg\,x_1$, where $x_1 = s_1\cos\theta$ is the horizontal projection. On the level stretch $x_2$ it does $\mu mg\,x_2$. So:

\[ mgh = \mu mg\,(x_1 + x_2) = \mu mg\,d, \] where $d$ is the total horizontal distance from start to stop.

Step 2: Coefficient

\[\mu = \frac{h}{d} = \frac{6}{25} = 0.24\]

$h = 6\ \mathrm{m}$ $25\ \mathrm{m} \ \text{(horizontal, start to stop)}$ $mgh = \mu mg \cdot (\text{horizontal distance})$ $\mu = \frac{6}{25} = 0.24$

On the incline, friction $\mu mg\cos\theta$ acts over a length whose horizontal projection is $s\cos\theta$, so friction work depends only on the horizontal distance covered. The whole path acts like one horizontal stretch.

Result

\[\boxed{\,\mu = \frac{h}{d} = \frac{6}{25} = 0.24\,}\]

Remark

About the original solution. It set the incline length to $\sqrt{25^2 + 6^2}$, i.e. took 25 m as the incline's horizontal length, and then added another 25 m of level path; that would give $\mu = 6/50 = 0.12$, and its stated 0.21 does not follow from its own equation. The distances in the problem only make sense if 25 m is the whole horizontal distance from the start to the stopping point, which gives $\mu = 0.24$. (If 25 m were measured from the foot of the incline, the incline's length would also have to be given.)