A 1.00 -kg block is pushed up and down a rough plank of length \(L=2.00 \mathrm{~m},\) inclined at \(30.0^{\circ}\) above the horizontal. From the bottom, it is pushed a distance \(L / 2\) up the plank, then pushed back down a distance \(L / 4,\) and finally pushed back up the plank until it reaches the top end. If the coefficient of kinetic friction between the block and plank is \(0.300,\) determine the work done by the block against friction.

The frictional force \(F_f\) acting on the block is given by the formula \(F_f = μN\), where \(μ\) is the coefficient of friction and \(N\) is the normal force acting on the block. In this case, μ is given as \(0.300\). Now, to calculate the normal force, we need to consider the weight of the block, which is \(mg\) (where m is the mass and g is the gravitational acceleration). Since the plank is inclined at an angle of \(30.0^{\circ}\), the normal force component will be \(mg\cos(30.0^{\circ})\). Then, we have: \(F_f = μmg\cos(30.0^{\circ}) = 0.300 \times 1.00~\text{kg} \times 9.81~\text{m/s^2} \times \cos(30.0^{\circ})\) Calculating this, we find that the frictional force is approximately \(F_f = 2.54~\text{N}\).

We have 3 parts of motion to consider: moving up L/2, moving down L/4, and moving up L/4. In each part, the work is given by the formula \(W = F \cdot d \cdot \cos(\theta)\). Since the force and distance are always either in the same direction (when moving up) or opposite directions (when moving down), the angle θ will be either \(0^{\circ}\) or \(180^{\circ}\). This simplifies the cosine term to either 1 (when moving up) or -1 (when moving down). Part 1 - Moving up L/2: \(W_{1} = F_f \cdot \frac{L}{2} \cdot \cos(0^{\circ}) = 2.54~\text{N} \cdot 1.00~\text{m} \cdot 1 = 2.54~\text{J}\) Part 2 - Moving down L/4: \(W_{2} = F_f \cdot \frac{L}{4} \cdot \cos(180^{\circ}) = 2.54~\text{N} \cdot 0.50~\text{m} \cdot (-1) = -1.27~\text{J}\) Part 3 - Moving up L/4: \(W_{3} = F_f \cdot \frac{L}{4} \cdot \cos(0^{\circ}) = 2.54~\text{N} \cdot 0.50~\text{m} \cdot 1 = 1.27~\text{J}\)

Now, we just need to sum up the work done against friction in each part of the motion to find the total work: \(W_{\text{total}} = W_{1} + W_{2} + W_{3} = 2.54~\text{J} + (-1.27~\text{J}) + 1.27~\text{J} = 2.54~\text{J}\) The total work done by the block against friction is \(2.54~\text{J}\).