$2.3.11.$ A body of mass $m$ starts to slide down from the upper end of a board of length $l$ that forms an angle $\alpha$ with the vertical. What kinetic energy will it acquire upon reaching the lower end of the board? Consider the case with no friction and the case when the coefficient of friction between the body and the board is $\mu < \cot\alpha$.
Solution
According to the work-energy theorem, the kinetic energy $E_k$ acquired by the body is equal to the sum of the works done by all forces acting on it: gravity, normal reaction force, and friction force. $$E_k = A_{mg} + A_N + A_{\text{fr}}$$
Since the board forms an angle $\alpha$ with the vertical, the vertical height the body drops is $h = l \cos \alpha$. The work done by gravity is: $$A_{mg} = mgh = mgl \cos \alpha$$ The work done by the normal force is zero ($A_N = 0$) because it is perpendicular to the displacement.
Case 1: No friction ($\mu = 0$) If there is no friction, the total kinetic energy is simply the work done by gravity: $$K = A_{mg} = mgl \cos \alpha$$
Case 2: With friction ($\mu < \cot\alpha$) Because the angle $\alpha$ is with the vertical, the normal force is $N = mg \sin \alpha$. The friction force is $F_{\text{fr}} = \mu N = \mu mg \sin \alpha$. The work done by the friction force (which is directed opposite to the displacement) is negative: $$A_{\text{fr}} = -F_{\text{fr}} l = -\mu mgl \sin \alpha$$
The total kinetic energy at the bottom of the board is: $$K_0 = mgl \cos \alpha - \mu mgl \sin \alpha = mgl(\cos \alpha - \mu \sin \alpha)$$