Answer:
Thus, the speed at B = 13 m/s
Explanation:
From the attached image:
Using the conservation of energy;
Iinital M.E = Final M.E
where;
M.E = mechanical energy
∴
[tex]\dfrac{1}{2} \times m \times v_A^2+ mgh_A = \dfrac{1}{2} \times m \times v_B^2+ mgh_B[/tex]
[tex]\dfrac{1}{2} \times m \times 3^2+m \times 9.8 \times 15 = \dfrac{1}{2} \times m \times v_B^2+ m\times 9.8 \times 7 \\ \\ \dfrac{1}{2} \times 3^2 + 9.8 \times 15 = \dfrac{1}{2} \times v^2 + 9.8 \times 7 \\ \\ \implies 151.5 = 0.5 v^2 + 68.6 \\ \\ 0.5v^2 = 151.5 - 68.6 \\ \\ 0.5 v^2 = 82.9 \\ \\ v^2 = \dfrac{82.9}{0.5} \\ \\ v^2 = 165.8 \\ \\ v = \sqrt{165.8} \\ \\ v = 12.9 \ m/s\\ \\ \mathbf{v \simeq 13 \ m/s}[/tex]
