While a roofer is working on a roof that slants at 36° above the horizontal, he accidentally nudges his 85.0-N toolbox, causing it to start sliding downward from rest. If it starts 4.25 m from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is 22.0 N?

The kinetic friction of force on the roof is f = 22 N

The length of the roof is: l = 4.25 m

The slope of the roof from horizontal is: $\theta =36°$

The weight of the toolbox is w = 85 N

The kinetic energy of the toolbox at the top edge of the inclined roof gets changed at the bottom edge of the inclined roof. This change occurs due to the frictional work of friction force opposite to the movement of the toolbox.

Apply the work-energy theorem to calculate the speed of the toolbox at the bottom edge of the inclined roof:

$\frac{1}{2}\left(\frac{w}{g}\right)v^2=(w\sin \theta -f)\mathrm{l}$

Here, g is the gravitational acceleration and its value is $9.8\mathrm{m}/{\mathrm{s}}^2$, $\mathrm{\theta }$is the angle of the inclined roof. v is the initial speed of the toolbox and l is the length of the inclined roof, w is the weight of the toolbox.

Therefore, the speed of the skier at the bottom of the hill is 8.16 m/s .

Substitute all the values in the above equation and we get,

$$\begin{gathered} \frac{1}{2}\left(\frac{(85\hspace{0.33em}N)}{9.8m/s^2}\right)v^2=\left[\right(85\hspace{0.33em}N\left)\right(\sin 36°)-22\hspace{0.33em}N](4.25\hspace{0.33em}m) \\ v=5.23m/s \end{gathered}$$

Therefore, the speed of the toolbox at the bottom edge of the inclined roof is 5.3 m/s .