A particle slides down along an inclined plane $A C$. If the plane is frictionless, kinetic energy of…
A particle slides down along an inclined plane $A C$. If the plane is frictionless, kinetic energy of particle at $C$ is
$m g y$
$m g x$
$m g\left(\frac{y}{\sin \theta}\right)$
$m g\left(\frac{y}{\cos \theta}\right)$
Solution
Let the mass of the particle be $m$. If the particle slides from height $y$ under the action of gravity, the velocity acquired by the particle is given by, $v=\sqrt{2 g y}$
Hence, kinetic energy, $\mathrm{KE}=\frac{1}{2} m v^2$
$
\begin{aligned}
& \Rightarrow \quad \mathrm{KE}=\frac{1}{2} m(\sqrt{2 g y})^2 \\
& =m g y \\
&
\end{aligned}
$
