If the coefficient of kinetic friction ($\mu_k$) is given, we would follow these steps to calculate the work done against friction and then the total work done. Step 1: Calculate the normal force ($F_N$). On an incline, the normal force is the component of the gravitational force perpendicular to the incline. $$F_N = m g \cos(\theta)$$ Where: $m$ is the mass of the crate (50 kg) $g$ is the acceleration due to gravity (9.8 m/s²) $\theta$ is the angle of incline (60°) Step 2: Calculate the force of kinetic friction ($F_f$). The force of kinetic friction is given by the product of the coefficient of kinetic friction and the normal force. $$F_f = \mu_k F_N$$ $$F_f = \mu_k (m g \cos(\theta))$$ Step 3: Calculate the distance along the incline ($d$). As calculated previously: $$d = \frac{h}{\sin(\theta)}$$ Where: $h$ is the height of the incline (20 m) $\theta$ is the angle of incline (60°) Step 4: Calculate the work done against friction ($W_f$). The work done against friction is the force of friction multiplied by the distance along the incline. $$W_f = F_f \times d$$ $$W_f = (\mu_k m g \cos(\theta)) \times \left(\frac{h}{\sin(\theta)}\right)$$ This can also be written as: $$W_f = \mu_k m g h \frac{\cos(\theta)}{\sin(\theta)}$$ $$W_f = \mu_k m g h \cot(\theta)$$ Step 5: Calculate the total work done ($W_{total}$). The total work done by the applied force to move the crate up the incline is the sum of the work done against gravity and the work done against friction. $$W_{total} = W_{gravity} + W_f$$ We know $W_{gravity} = m g h$. So, the formula for total work done would be: $$W_{total} = m g h + \mu_k m g h \cot(\theta)$$ $$W_{total} = m g h (1 + \mu_k \cot(\theta))$$ This formula would allow you to calculate the total work done if the coefficient of kinetic friction ($\mu_k$) were provided.