## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

We can find the maximum possible force that the force of static friction can provide for the 3.0-kg block without the 4.0-kg block slipping. Let $m_u$ be the mass of the 4.0-kg block. $F_f = m_u~g~\mu_s$ $F_f = (4.0~kg)(9.80~m/s^2)(0.60)$ $F_f = 23.52~N$ We can find the acceleration of the bottom block when this force of static friction is provided by the contact between the two blocks. Let $m_b$ be the mass of the 3.0-kg block. $\sum F = m_b~a$ $F_f - (m_b+m_u)~g~\mu_k = m_b~a$ $a = \frac{F_f - (m_b+m_u)~g~\mu_k}{m_b}$ $a = \frac{(23.52~N) - (3.0~kg+4.0~kg)(9.80~m/s^2)(0.20)}{3.0~kg}$ $a = 3.27~m/s^2$ Since both blocks move together, this will be the acceleration of the system of both blocks. We can find the time it takes to travel 5.0 meters. $d = \frac{1}{2}at^2$ $t = \sqrt{\frac{2d}{a}}$ $t = \sqrt{\frac{(2)(5.0~m)}{3.27~m/s^2}}$ $t = 1.75~s$ The least amount of time to travel 5.0 meters without the top block slipping is 1.75 seconds.