How much work do you do as you exert a \(75-\mathrm{N}\) force to push a shopping cart through a 12 -m-long supermarket aisle?

Understanding how to calculate work in physics is essential for comprehending energy transfer during motion. Work, represented by the symbol 'W', is done when a force 'F' causes an object to move through a distance 'd'. It's crucial to note that the force must have a component in the direction of the object's displacement for work to occur.

Mathematically, work is expressed by the equation: \[ W = F \times d \times \text{cos}(\theta) \], where \( \theta \) represents the angle between the force applied and the displacement of the object. In the case of pushing a shopping cart with a constant force in a straight line, you would simply multiply the force exerted by the distance to calculate the total work done, providing the textbook example of 'work' in a physics context.

The concepts of force and displacement are central to calculating work in physics. A force refers to any interaction that, when unopposed, will change the motion of an object. In more technical terms, it's the push or pull acting upon an object resulting from the object's interaction with another object. Displacement, on the other hand, signifies the object's change in position.

To result in work, the force must be exerted in a direction that causes displacement. This means that if you apply force to an immovable wall, no work is done since there is no displacement. However, with our supermarket aisle example, the force exerted results in the cart moving through the aisle, indicating work is being performed. The actual calculation of work done takes both the magnitude of the force and the distance over which it is exerted into account.

The cosine of the angle between the force applied and the direction of displacement plays a vital role in the calculation of work. The use of cosine in the work formula \[ W = F \times d \times \text{cos}(\theta) \] adjusts the amount of work done based on this angle. This factor ensures that only the component of the force acting in the direction of motion contributes to the work done.

For instance, if the force is exerted perpendicular to the direction of motion (\( \theta = 90^\circ \)), the cosine of 90 degrees is 0, which results in zero work done in the direction of displacement. Conversely, when the angle is 0 degrees, as in our textbook problem, the cosine of 0 degrees is 1. This signifies that the full magnitude of the force contributes to the work done, leading to the simplified work calculation for this scenario.