## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

We can find the speed of the pulse. $v = \frac{d}{t}$ $v = \frac{4.0~m}{24.0\times 10^{-3}~s}$ $v = 166.7~m/s$ Note that the mass of the middle part of the wire is 30.0 g. We can find the tension $T_m$ in the middle part of the wire. $\sqrt{\frac{T_m}{\mu}} = v$ $T_m = v^2~\mu$ $T_m = v^2~(\frac{m}{L})$ $T_m = (166.7~m/s)^2~(\frac{0.0300~kg}{4.0~m})$ $T_m = 208.4~N$ This tension is equal to the horizontal component of the tension $T_x$ in the left section of the wire. We can find the vertical component $T_y$ of the tension in the left section of the wire. $\frac{T_y}{T_x} = tan(40^{\circ})$ $T_y = T_x~tan(40^{\circ})$ $T_y = (208.4~N)~tan(40^{\circ})$ $T_y = 174.9~N$ The vertical component of the tension is equal to the weight of the block. We can find the mass of the block. $mg = T_y$ $m = \frac{T_y}{g}$ $m = \frac{174.9~N}{9.80~m/s^2}$ $m = 17.8~kg$ The mass of each block is 17.8 kg.