#### Answer

0.225

#### Work Step by Step

Here we’ll use the equation for the change in thermal energy, which we had previously found:
$\Delta E_{th} = f_{k}d = μ_{k}F_{N}d$
Solving for $μ_{k}$, we get:
$μ_{k} = \frac{\Delta E_{th}}{F_{N}d}$
In Problem 45c, we found that $\Delta E_{th}$ = 30.1 J, and we also know that d = 4.06 m. So all we need to find is the normal force with Newton’s second law for vertical components. Since the block is not moving vertically, $F_{net, y}$ = 0.
The forces acting vertically are the gravitational force, the inclined applied force and the normal force, so:
0 = $F_{g} + F_{app,y} + F_{N}$
0 = $–mg + F sin\theta + F_{N}$
$F_{N} = mg – F_{app} sin\theta$
Plugging all this in we get:
$μ_{k} = \frac{\Delta E_{th}}{F_{N}d}$
$μ_{k} = \frac{\Delta E_{th}}{(mg – F_{app} sin\theta)d}$
$μ_{k} = \frac{30.1 J}{((3.57 kg)(9.8 m/s²) – (7.68 N) sin 15^{\circ})(4.06 m)}$
= 0.225