Question

a uniform rod of mass 2kg and length 1m is lying on a horizontal surface. if the work done in raising one end of the rod through an angle 45⁰ is 'W', the work done in raising it further through 45⁰ is?

Answer 1

When the the rod is raised through an angle 45 degrees, the CoM of the rod moves up by $=0.5\times \sin {45}^0=\frac{0.5}{\sqrt{2}}=\frac{1}{2\sqrt{2}}m$
Work done, $W=2\times 10\times \frac{1}{2\sqrt{2}}=5\sqrt{2}J$
The CoM will raise again by $\left[\frac{1}{2}-\frac{1}{2\sqrt{2}}\right]m$
Work done in this case,
$$\begin{gathered} W_1=2\times 10\times \left(\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)=10\times \left(1-\frac{1}{\sqrt{2}}\right)=\left(\sqrt{2}-1\right)5\sqrt{2}J \\ {W}_1=\left(\sqrt{2}-1\right)W \end{gathered}$$