Chapter 8 - Potential Energy and Conservation of Energy - Problems - Page 205: 38a

$8.37m/s$

Finding the kinetic energy at point A: $E_A=E_B$ $K_A + U_A = K_B + U_B$ $K_A = K_B + U_B - U_A$ $K_A = 12J + 4J - 9J$ $K_A = 7J$ Using $K_A$ to find the speed at $x=3.5m/s^2$ $K_A = \frac{mv^2}{2}$ $2K_A = {mv^2}$ $\frac{2K_A}{m} = {v^2}$ $v =\sqrt \frac{2K_A}{m} $ $v =\sqrt \frac{2 \times 7J}{0.2kg} $ $v = 8.37m/s$