Answer
The work done by the executive on the briefcase is $15.6~J$
Work Step by Step
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 $15.6~J$
