Physics - W.E.P. Question with Solution | TestHub

According to the work-kinetic energy theorem, the total work done on the block equals the change in its kinetic energy. The work done by gravity is $W_g=mgL\sin \theta$. The normal force on the inclined plane is $N=mg\cos \theta$, so the work done by friction is $W_f=-{\mu }_{k}NL=-{\mu }_{k}mgL\cos \theta$. Since the block starts from rest, its initial kinetic energy is $0$. If its final speed is $v$, the final kinetic energy is $\frac{1}{2}m{v}^2$. So, $W_g+W_f=\frac{1}{2}m{v}^2-0$. This gives $mgL\sin \theta -{\mu }_{k}mgL\cos \theta =\frac{1}{2}m{v}^2$. Solving for $v$, we get $v=\sqrt{2gL(\sin \theta -{\mu }_k\cos \theta )}$.