91Ó°ÊÓ

When discussing forces acting on an object, work can be calculated using the formula: \(W = F \cdot d \cdot \cos(\theta)\), where \(W\) is the work done, \(F\) is the force, \(d\) is the distance, and \(\theta\) is the angle between the force and direction of motion. For instance, in this exercise, the worker's force is \(209 \, \mathrm{N}\), the distance the crate moves is \(1.50 \, \mathrm{m}\), and the angle is \(0^{\circ}\) since the force is parallel to the incline. Hence, the work done by the worker is \(209 \, \mathrm{N} \times 1.50 \, \mathrm{m} \times \cos(0^{\circ}) = 313.5 \, \mathrm{J}\). Understanding how to align forces and their respective angles is key to solving work-related problems.

Gravitational force acts vertically downward on an object. It's calculated using: \(F_g = m \cdot g\), where \(m\) is the mass and \(g\) is the acceleration due to gravity. In our case, the crate's mass \(m\) is \(25.0 \, \mathrm{kg}\) and \(g = 9.8 \, \mathrm{m/s^2}\). Thus, \(F_g = 25.0 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} = 245 \, \mathrm{N}\). Since the crate is on an incline, we need the component of gravitational force acting parallel to the incline. This is given by \(F_{g,\parallel} = F_g \cdot \sin(25^{\circ})\). Hence, \(245 \, \mathrm{N} \times \sin(25^{\circ}) \approx 103.5 \, \mathrm{N}\). The work done by this component, acting in opposite direction to crate movement, is \(-103.5 \, \mathrm{N} \times 1.50 \, \mathrm{m} = -155.3 \, \mathrm{J}\).

The normal force is perpendicular to the surface on which the object rests. In this problem, it acts perpendicular to the incline. It's vital to understand that since normal force acts at a \(90^{\circ}\) angle to the direction of motion, the work done by it is zero. This is because \(\cos(90^{\circ}) = 0\). Therefore, even though the normal force is present, it doesn't contribute to the work as per the formula \(W_{normal} = F_{normal} \cdot d \cdot \cos(90^{\circ})\). So, \(W_{normal} = 0 \, \mathrm{J}\).

To find the total work done on the crate, sum the work done by all individual forces. Here, we have the work by the worker (\(W_{worker} = 313.5 \, \mathrm{J}\)), the gravitational force (\(W_{gravity} = -155.3 \, \mathrm{J}\)), and the normal force (\(W_{normal} = 0 \, \mathrm{J}\)). Adding these gives: \(W_{total} = 313.5 \, \mathrm{J} + (-155.3 \, \mathrm{J}) + 0 \, \mathrm{J} = 158.2 \, \mathrm{J}\). This total work gives a clear understanding of the energy changes that took place as the crate was moved up the incline.