Chapter 9 - Problems - Page 362: 45

(a) $a = 9.80~m/s^2$ upward (b) $a = 3.27~m/s^2$ upward (c) $a = 68.6~m/s^2$ upward

(a) We can express the density of water $\rho_w$ in terms of the density of balsa wood $\rho$: $\frac{\rho_w}{\rho} = \frac{1.0~g/cm^3}{0.50~g/cm^3}$ $\rho_w = 2.0~\rho$ According to Archimedes' principle, the buoyant force is equal to the weight of the water that is displaced. Let $V$ be the volume of the balsa wood. We can find the initial acceleration of the balsa wood: $\sum F = ma$ $F_b-mg = ma$ $\rho_w~V~g-\rho~V~g = ma$ $2.0~\rho~V~g-\rho~V~g = ma$ $\rho~V~g = ma$ $mg = ma$ $a = g$ $a = 9.80~m/s^2$ upward (b) We can express the density of water $\rho_w$ in terms of the density of maple $\rho$: $\frac{\rho_w}{\rho} = \frac{1.0~g/cm^3}{0.750~g/cm^3}$ $\rho_w = \frac{4}{3}~\rho$ According to Archimedes' principle, the buoyant force is equal to the weight of the water that is displaced. Let $V$ be the volume of the maple. We can find the initial acceleration of the maple: $\sum F = ma$ $F_b-mg = ma$ $\rho_w~V~g-\rho~V~g = ma$ $\frac{4}{3}~\rho~V~g-\rho~V~g = ma$ $\frac{1}{3}~\rho~V~g = ma$ $\frac{1}{3}~mg = ma$ $a = \frac{g}{3}$ $a = \frac{9.80~m/s^2}{3}$ $a = 3.27~m/s^2$ upward (c) We can express the density of water $\rho_w$ in terms of the density of the ping-pong ball $\rho$: $\frac{\rho_w}{\rho} = \frac{1.0~g/cm^3}{0.125~g/cm^3}$ $\rho_w = 8~\rho$ According to Archimedes' principle, the buoyant force is equal to the weight of the water that is displaced. Let $V$ be the volume of the ping-pong ball. We can find the initial acceleration of the ping-pong ball: $\sum F = ma$ $F_b-mg = ma$ $\rho_w~V~g-\rho~V~g = ma$ $8~\rho~V~g-\rho~V~g = ma$ $7~\rho~V~g = ma$ $7~mg = ma$ $a = 7~g$ $a = (7)(9.80~m/s^2)$ $a = 68.6~m/s^2$ upward