## Essential University Physics: Volume 1 (3rd Edition)

$\mu=.46$
We must first find an expression for the acceleration, which is equal to the force of gravity causing the boxes to move down the ramp minus the force of friction, which is all divided by the mass of the box. Thus, we find: $a = \frac{mgsin\theta-F_n \mu}{m} \\ a = \frac{mgsin\theta-mgcos\theta \mu}{m}\\ a = gsin\theta-gcos\theta \mu \\ a = gsin30-gcos30 \mu \\ a = 4.905-8.5\mu$ Thus, we find: $\Delta x = v_0t + \frac{1}{2}at^2 \\ 5.4 = 0t+\frac{1}{2}( 4.905-8.5\mu)(3.3^2)\\ \mu=.46$