Essential University Physics: Volume 1 (3rd Edition)

$25.33 \ seconds$
We know that the overall mass of the system is: $=73+940=1013 \ kg$ We know that the acceleration is equal to the force of gravity minus the force of friction over the overall mass. Thus, we find: $a= \frac{(73)(9.81)-(.057)(940)(9.81)}{1013}=.159\ m/s^2$ We now can find the change in time, using the same equation that the book uses in example 5.4. Doing this, we find: $t =\sqrt{\frac{2x}{a}}=\sqrt{\frac{2(51)}{.159}}=25.33 \ seconds$