The first step is to find the total potential energy at 45 degrees:

To do this you must find the height using trigonometry

\(\displaystyle{P}{E}={m}{g}{h}={\left({1.5}\right)}\cdot{\left({9.81}\right)}\cdot{\left({200}-{\sin{{\left({45}\right)}}}\cdot{\left({\frac{{{200}}}{{{1000}}}}\right)}={0.86198}\ {J}\right.}\)

Since energy is conserved the kinetic energy at the bottom of thependulum must equal the potential at the top minus the energydissipated.

\(\displaystyle{K}{E}={0.5}\cdot{m}\cdot{v}^{{2}}={0.5}\cdot{\left({1.5}\right)}\cdot{\left({\frac{{{200}}}{{{1000}}}}\cdot{5.2}\right)}^{{2}}={0.8112}\ {J}\)

So energy dissipated =PE-KE=0.05078 J

To do this you must find the height using trigonometry

\(\displaystyle{P}{E}={m}{g}{h}={\left({1.5}\right)}\cdot{\left({9.81}\right)}\cdot{\left({200}-{\sin{{\left({45}\right)}}}\cdot{\left({\frac{{{200}}}{{{1000}}}}\right)}={0.86198}\ {J}\right.}\)

Since energy is conserved the kinetic energy at the bottom of thependulum must equal the potential at the top minus the energydissipated.

\(\displaystyle{K}{E}={0.5}\cdot{m}\cdot{v}^{{2}}={0.5}\cdot{\left({1.5}\right)}\cdot{\left({\frac{{{200}}}{{{1000}}}}\cdot{5.2}\right)}^{{2}}={0.8112}\ {J}\)

So energy dissipated =PE-KE=0.05078 J