Question

5A pendulum of mass 200 g is pulled horizontally sideways by 30 cm. The length of the pendulum string is 50 cm.aCalculate the potential energy of the pendulum in this position. Assume g = 9.8 N/kg

Answer 1

Answer:

Approximately $0.196\; \rm J$.

Explanation:

Convert all units of measure to standard units:

Length of the pendulum string:

$\begin{aligned} 50\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm} = 0.50\; \rm m\end{aligned}$.

Horizontal distance between the pendulum and the fixed point:

$\begin{aligned} 30\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm} = 0.30\; \rm m\end{aligned}$.

Mass of the pendulum:

$\begin{aligned}200\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} = 0.200\; \rm g \end{aligned}$

Refer to the diagram attached. Apply the Pythagorean Theorem to find the current vertical distance between the fixed point and the pendulum:

$\begin{aligned}\sqrt{(0.50\; \rm m)^{2} - (0.30\; \rm m)^{2}} &= 0.40\; \rm m\end{aligned}$.

Thus, the current position of the pendulum is $h = 0.50\; \rm m - 0.40\; \rm m = 0.10\; \rm m$ above its equilibrium position.

Given that $m = 0.200\; \rm kg$ and $g = 9.8\; \rm N \cdot kg^{-1}$, calculate the gain in the gravitational potential energy of this pendulum:

$\begin{aligned}\text{GPE} &= m \cdot g \cdot h \\ &= 0.200\; \rm kg \times 9.8\; \rm N \cdot kg^{-1} \times 0.10\; \rm m \\ &= 0.196\; \rm J\end{aligned}$.

Thus, the gravitational potential energy of this pendulum (compared to the same pendulum at the equilibrium position) would be $0.196\; \rm J$.