Usaha dan Energi Pembahasan Soal 1

L cos 60 = 2,5 cm × 0,5 = 1,25 cm

Beda tinggi antara posisi (1) dan (2) = L - L cos 60 = 2,5 cm - 1,25 cm = 1,25 cm = 0,0125 m

Hukum Kekekalan Energi

\begin{equation*}
\begin{split}
EM_1 & = EM_2 \\\\
EP_1 + EK_1 & = EP_2 + EK_2 \\\\
m \:.\: g \:.\: h_1 + 0 & = 0 + \tfrac{1}{2} \:.\: m \:.\: v_2^2 \\\\
\cancel {m} \:.\: g \:.\: h_1 & = \tfrac{1}{2} \:.\: \cancel {m} \:.\: v_2^2 \\\\
g \:.\: h_1 & = \tfrac{1}{2} \:.\: v_2^2 \\\\
2 \:.\: g \:.\: h_1 & = v_2^2 \\\\
v_2 & = \sqrt{2 \:.\: g \:.\: h_1} \\\\
v_2 & = \sqrt{2 \:.\: 10 \:.\: 0,0125} \\\\
v_2 & = \sqrt{0,25} \\\\
v_2 & = 0,5 \text{ m/s}
\end{split}
\end{equation*}