Question

Explain the work-energy theorem, including the direction energy moves and how it relates to positive and negative work.

Answer 1

According to this theorem, when an object slows down, its final kinetic energy is less than its initial kinetic energy, the change in its kinetic energy is negative, and so is the net work done on it. If an object speeds up, the net work done on it is positive.

Answer 2

The work-energy theorem, including the direction energy moves and how it relates to positive and negative work are:

• This work-energy theorem suggests that all forces acting on an object generate the same amount of work, resulting in the same change in the particle's kinetic energy. Through describing the work of a torque plus rotational kinetic energy, the concept could be generalized to rigid - body.
• Kinetic Energy: On the block, a force is at work. The amount of work causes the block's kinetic energy to grow. This work-energy theorem generalises this relationship.
• The change in kinetic energy KE is equal to the work W done by that of the net force on a particle:

      W=ΔKE=$\frac{1}{2}$ m$v^{2}$ f - $\frac{1}{2}$ m$v^{2}$i

      where $v^{f}$ and $v^{i}$ are the particle's initial and final speeds, respectively, and m is the particle's mass.

• For the purpose of simplicity, we'll assume that the resultant force F is constant in size and direction, and that it's parallel to the particle's motion.
• The particle is travelling in a single direction at constant acceleration. The expression F = ma (Newton's second law) describes the link between net force and acceleration, and the particle's displaced d can be calculated using the equation:

           $v^{2}$f = $v^{2}$i + 2ad

          acquiring,

          d=$\frac{v^{2}fv^{2}i }{2a }$

• The product of the net force's magnitude (F=ma) and the particle's displacement is used to determine the net force's work. When the aforementioned equations are substituted, we get:

          W=fd=ma$\frac{v^{2}fv^{2}i }{2a }$ =$\frac{1}{2}$ m$v^{2}$ f - $\frac{1}{2}$ m$v^{2}$i= K$E_{f}$ - K$E_{i}$ =ΔKE