Answer: The work-energy theorem states that a force acting on a particle as it moves over a DISTANCE changes the KINETIC energy of the particle if the force has a component parallel to the motion.
Explanation:
The work- energy theorem states that the work done when forces act on particles as it moves over a distance is equal to the change in Kinetic energy of the particle if the force has a component parallel to the motion. To express this definition an equation is used:
W= ∆K.E =(final K. E. - initial K. E)
Therefore W= ½mv²(final) - ½mv²(initial)
Where W is work done by net force and
K.E ( final minus the initial) is the particle change in Kinetic Energy.
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. When calculating the net work, you must include all the forces that act on an object.
This theorem also shows that since work done on a particle can lead to change in its kinetic energy, it implies that work can transfer energy from one form to another.
Answer:
The work-energy theorem states that a force acting on a particle as it moves over a distance changes the kinetic energy of the particle if the force has a component parallel to the motion.
Explanation:
The correct answer is presented below and all reasons are presented to explain all facts:
The work-energy theorem states that a force acting on a particle as it moves over a distance changes the kinetic energy of the particle if the force has a component parallel to the motion.
Reasons:
According to the Work-Energy Theorem, the work done on a particle ([tex]W[/tex]) equals the change in its kinetic energy ([tex]\Delta K[/tex]). That is:
[tex]W = \Delta K[/tex] (1)
By definition of work we expand this definition:
[tex]\oint \vec F\,\bullet\,d\vec s = \Delta K[/tex] (2)
Where:
[tex]\vec F[/tex] - Vector force.
[tex]\vec s[/tex] - Vector travelled distance.
And by definition of dot product we conclude that:
[tex]\int\limits_{A}^{B}{\|\vec F\|\cdot \|d\vec{s}\|\cdot \cos \phi} = \Delta K[/tex]
Where:
[tex]\|\vec F\|[/tex] - Magnitude of the vector force.
[tex]\|d\vec s\|[/tex] - Magnitude of the differential of the vector travelled distance.
[tex]\phi[/tex] - Angle between vectors, measured in sexagesimal degrees.
[tex]A, B[/tex] - Initial and final position of the particle.
From this expression we infer that change in kinetic energy is maximum if and only if [tex]\phi = 0^{\circ}[/tex] in every point of the path travelled by the particle. In addition, change in kinetic energy occurs when component of force parallel to path is not zero.
