Answer
30.1 J
Work Step by Step
Here we can use the definition of work done on a system:
$W = \Delta E = \Delta E_{mec} + \Delta E_{th} + \Delta E_{int} = \Delta K + \Delta U + \Delta E_{th} + \Delta E_{int}$
Since the block is moving horizontally and at constant speed, both $\Delta U$ = 0 and $\Delta K$ = 0. The only change in internal energy being discussed is $\Delta E_{th}$, so we can assume that any other $\Delta E_{int}$ is zero. That leaves us with:
$W = \Delta E_{th}$
In Problem 45a, we calculated that W = 30.1 J. Therefore,
$\Delta E_{th}$ = 30.1 J
