Answer
At the turning point, the speed of particle is zero. Let the position of the right turning point be $x_{R} .$ From the figure shown in below, we find $x_{R}$ to be
$$
\frac{16.00 \mathrm{J}-0}{x_{R}-7.00 \mathrm{m}}$$$$=\frac{24.00 \mathrm{J}-16.00 \mathrm{J}}{8.00 \mathrm{m}-x_{R}} \Rightarrow x_{R}$$$$=7.67 \mathrm{m} .
$$
Work Step by Step
At the turning point, the speed of particle is zero. Let the position of the right turning point be $x_{R} .$ From the figure shown in below, we find $x_{R}$ to be
$$
\frac{16.00 \mathrm{J}-0}{x_{R}-7.00 \mathrm{m}}$$$$=\frac{24.00 \mathrm{J}-16.00 \mathrm{J}}{8.00 \mathrm{m}-x_{R}} \Rightarrow x_{R}$$$$=7.67 \mathrm{m} .
$$
