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

$m_1 = 0.080~kg$ $m_3 = 0.50~kg$
Let $F$ be the force provided by one rubber band. We can find the value of $F$ using graph 2. So, $5F = (m_2)(5~m/s^2)$ $F = \frac{(0.20~kg)(5~m/s^2)}{5}$ $F = 0.20~N$ From graph 1, we can see that $m_1$ has an acceleration of $5~m/s^2$ when there are two rubber bands. We can use graph 1 to set up an equation for $m_1$. So, $2F = (m_1)(5~m/s^2)$ $m_1 = \frac{2F}{5~m/s^2}$ $m_1 = \frac{(2)(0.20~N)}{5~m/s^2}$ $m_1 = 0.080~kg$ From graph 3, we can see that $m_3$ has an acceleration of $2~m/s^2$ when there are five rubber bands. We can use graph 3 to set up an equation for $m_3$. So, $5F = (m_3)(2~m/s^2)$ $m_3 = \frac{5F}{2~m/s^2}$ $m_3 = \frac{(5)(0.20~N)}{2~m/s^2}$ $m_3 = 0.50~kg$