Assuming they give us coefficient of friction, what formula do we use?

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°)

The force of kinetic friction is given by the product of the coefficient of kinetic friction and the normal force.

$$F_f = \mu_k F_N$$ $$F_f = \mu_k (m g \cos(\theta))$$

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°)

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:

$$W_f = \mu_k m g h \frac{\cos(\theta)}{\sin(\theta)}$$ $$W_f = \mu_k m g h \cot(\theta)$$

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.