A sled starts from rest at point A of a slope, slides down…

A sled starts from rest at point A of a slope, slides down to its foot B and continues along a horizontal stretch to C, where it stops. $AB = BC$, and the coefficient of friction $\mu = 0.1$ is the same along the whole path. Find the angle of the slope.

Explanation of Method

Slope angle from the sled's stopping point.

Step 1: Energy balance

With $AB = BC = l$: on the slope friction is $\mu mg\cos\theta$, on the level stretch it is $\mu mg$: $mgl\sin\theta = \mu mg\cos\theta \cdot l + \mu mg \cdot l$.

\[ \sin\theta = \mu\,(1 + \cos\theta). \]

Step 2: Solve

Using $\sin\theta = 2\sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2}$ and $1 + \cos\theta = 2\cos^2\tfrac{\theta}{2}$: $\tan\frac{\theta}{2} = \mu \;\Longrightarrow\; \theta = 2\arctan 0.1$.

\[\theta \approx 11.4^\circ\]

$A$ $B$ $C$ $\theta$ $l$ $l$ $\sin\theta = \mu(1 + \cos\theta) \Rightarrow \tan\tfrac{\theta}{2} = \mu$

Drawn to scale for $\mu = 0.1$ ($\theta \approx 11.4^\circ$). On the slope friction is $\mu mg\cos\theta$, on the level stretch it is the larger $\mu mg$.

Result

\[\boxed{\,\theta \approx 11.4^\circ\,}\]

Remark

About the original solution. It used $\mu mg\cos\theta$ for the friction on the horizontal stretch as well, which gives $\tan\theta = 2\mu$ and 11.3°. On level ground the normal force is the full $mg$, so the correct condition is $\tan(\theta/2) = \mu$, $\theta \approx 11.4^\circ$. The numbers are close only because the angle is small.