A small object is sliding on a smooth horizontal floor along a vertical wall. The wall makes a smooth turn by angle ${\theta }_0$ . Coefficient of friction between the wall and the block is $\mu$. Speed of the object before the turn is u. find its speed (v) just after completing the turn. [The turn is smooth and there are no sharp corners]

Consider the object at point P on the curve. Let its velocity at this point be V making an angle $\theta$ with the original direction. Let the centre curvature of the path at this point be O and the radius of curvature be R.

The normal force is  $N=\frac{m{v}^2}{R}$
Kinetic friction force on the object is $f=\mu N=\frac{\mu m{v}^2}{R}$  work done by the friction in small angular displacement $d\theta$ of the block is $d{W}_f=-f\left(Rd\theta \right)=-\mu m{v}^{2}d\theta$ from work energy theorem, work done by the friction is equal to change in KE of the object
$K=\frac{1}{2}m{v}^2$
$dK=mvdv$
$\therefore mvdv=-\mu m{v}^{2}d\theta$
$\Rightarrow \frac{dv}{v}=-\mu d\theta \Rightarrow \underset{u}{\overset{v}{\int }}\frac{dv}{v}=-\mu \underset{0}{\overset{{\theta }_0}{\int }}d\theta$
$$\begin{gathered} \Rightarrow \ln v-\ln u=-\mu {\theta }_0\Rightarrow \ln \left(\frac{v}{u}\right)=-\mu {\theta }_0 \\ \Rightarrow \frac{v}{u}=e^{-\mu {\theta }_0} \\ \Rightarrow v=u{e}^{-\mu {\theta }_0} \end{gathered}$$
Answer does not shape of the curve.