7.11
Trabalho Realizado Sobre um Plano Inclinado
A estrutura conceitual do centro de massa ajuda a descrever facilmente o trabalho realizado em corpos rígidos. Já que as forças internas em um corpo r…
Considere uma pessoa esquiando em uma encosta. Supondo que a força devido ao atrito nos esquis seja insignificante sobre o gelo, um diagrama de corpo livre pode ser desenhado.
O componente da força gravitacional ao longo do deslocamento move a pessoa para frente. As forças perpendiculares ao deslocamento não têm papel no movimento e, portanto, não contribuem para o trabalho realizado.
Portanto, o trabalho realizado é simplesmente o componente de força paralela multiplicado pelo deslocamento.
Outra maneira de resolver isso é usando o vetor de força em vez de seu componente.
Aqui, o ângulo entre a força e o deslocamento é considerado. Observe que a expressão final para o trabalho realizado é a mesma.
No entanto, se o deslocamento do esquiador for na direção oposta da força, de modo que comece e termine em repouso, o trabalho realizado é negativo.
Se o ângulo do plano inclinado for zero, a força é perpendicular ao deslocamento, de modo que o trabalho realizado pela força gravitacional sobre a pessoa torna-se zero.
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Q1: Why do only force components parallel to displacement contribute to work on an inclined plane?
Forces perpendicular to displacement do not contribute to work because work depends on the component of force acting in the direction of motion. On an inclined plane, only the gravitational force component along the slope moves the object forward. The normal force, perpendicular to the plane, does no work regardless of the displacement distance.
Q2: How does the angle between force and displacement affect work calculation?
Work is calculated by multiplying the force magnitude by the displacement and the cosine of the angle between them. When the angle is zero, force and displacement align, maximizing work. When perpendicular (90 degrees), work becomes zero. On an inclined plane with zero slope angle, gravitational force becomes perpendicular to displacement, resulting in zero work.
Q3: When is work done by gravity negative on an inclined plane?
Work done by gravity is negative when displacement opposes the gravitational force direction, such as when an object moves up the slope against gravity. If the object starts and ends at rest while moving upward, negative work indicates energy is being stored or removed from the system, opposing gravity's natural downward pull.
Q4: What role does the center-of-mass framework play in calculating work on rigid bodies?
The center-of-mass framework simplifies work calculations for rigid bodies by allowing internal forces to be ignored, since they do no work. Only external forces like gravity need consideration. This approach applies equally to inclined planes, where only the gravity component along the plane contributes to work done on the body's center of mass.
Q5: Why is work done by gravity zero during horizontal motion?
During horizontal motion, gravitational force acts vertically downward while displacement occurs horizontally. Since force and displacement are perpendicular, their dot product is zero, resulting in zero work by gravity. Any change in kinetic energy during horizontal motion comes entirely from other forces like friction or applied forces, not gravity.
Q6: How do you calculate work done by gravity using the line integral method?
Work done by gravity is the line integral of gravitational force over the displacement path. The result is positive if gravity increases kinetic energy (object moving downward) and negative if work opposes gravity (object moving upward). On an inclined plane, only the gravity component parallel to the slope contributes to this line integral calculation.
Q7: What is the difference between using force components versus the angle method for work calculations?
Both methods yield identical results. The component method multiplies the parallel force component by displacement directly. The angle method multiplies total force magnitude by displacement by the cosine of the angle between them. On inclined planes, both approaches account for only the gravity component along the slope, producing the same work value.