## College Physics (4th Edition)

The work done by the executive on the briefcase is $15.6~J$
We can find the change in kinetic energy of the briefcase: $\Delta KE = \frac{1}{2}mv^2$ $\Delta KE = \frac{1}{2}(5.00~kg)(2.50~m/s)^2$ $\Delta KE = 15.6~J$ The work done on the briefcase is equal to the change in kinetic energy of the briefcase. Therefore, the work done by the executive on the briefcase is $15.6~J$