91Ó°ÊÓ

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 \(2.54~\text{J}\).