Answer:
Approximately [tex]0.196\; \rm J[/tex].
Explanation:
Convert all units of measure to standard units:
Length of the pendulum string:
[tex]\begin{aligned} 50\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm} = 0.50\; \rm m\end{aligned}[/tex].
Horizontal distance between the pendulum and the fixed point:
[tex]\begin{aligned} 30\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm} = 0.30\; \rm m\end{aligned}[/tex].
Mass of the pendulum:
[tex]\begin{aligned}200\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} = 0.200\; \rm g \end{aligned}[/tex]
Refer to the diagram attached. Apply the Pythagorean Theorem to find the current vertical distance between the fixed point and the pendulum:
[tex]\begin{aligned}\sqrt{(0.50\; \rm m)^{2} - (0.30\; \rm m)^{2}} &= 0.40\; \rm m\end{aligned}[/tex].
Thus, the current position of the pendulum is [tex]h = 0.50\; \rm m - 0.40\; \rm m = 0.10\; \rm m[/tex] above its equilibrium position.
Given that [tex]m = 0.200\; \rm kg[/tex] and [tex]g = 9.8\; \rm N \cdot kg^{-1}[/tex], calculate the gain in the gravitational potential energy of this pendulum:
[tex]\begin{aligned}\text{GPE} &= m \cdot g \cdot h \\ &= 0.200\; \rm kg \times 9.8\; \rm N \cdot kg^{-1} \times 0.10\; \rm m \\ &= 0.196\; \rm J\end{aligned}[/tex].
Thus, the gravitational potential energy of this pendulum (compared to the same pendulum at the equilibrium position) would be [tex]0.196\; \rm J[/tex].
