Dragging a Board

The board's motion begins in region 1 with the front tip of the board on the boundary between the two regions. By the end of the board's motion, the board now lies in region 2 with the back tip of the board on the boundary between the two regions. Let $t_1$ represent the amount of time the board spends in region 1. Let $t_2$ represent the amount of time the board spends in region 2. Let $T$ represent the total travel time of the board. We know that $t_1 + t_2 =T$ . In addition, because the velocity of the board remains constant throughout its motion, $t_1$ must be the same as $t_2$ . Therefore, $t_1 = t_2 =t$ and thus, $t=\frac{T}{2}$ . This means that on average, the board spends half on its time in region 1 and the other half in region 2. Thus, we can take the sum of the work done by both regions and average them as follows: $W =\frac{-μ_1 MgL - μ_2 MgL}{2}$
* Note that there are negative signs in front of each expression for work in the numerator. This is due to the fact that the force of friction acts in the OPPOSITE direction of the board's motion. Therefore, the angle between the force and the motion is 180° and $cos (180°) = -1$

Factoring out $MgL$ and simplifying yields: $W=-\frac{MgL}{2}$ $(μ_1+μ_2)$

Plug the given values for $L$ , $μ_1$ , $μ_2$ , $M$ and the constant $g$ into the equation above and solve. Solving yields -16 J for the net work done by friction.