Is Work The Change In Kinetic Energy

Kinetic Energy and Piece of work-Energy Theorem

The work-free energy theorem states that the work done by all forces acting on a particle equals the change in the particle'due south kinetic free energy.

Learning Objectives

Outline the derivation of the work-energy theorem

Central Takeaways

Cardinal Points

The piece of work W done by the internet force on a particle equals the modify in the particle'southward kinetic energy KE: [latex]\text{W}=\Delta \text{KE}=\frac{i}{2} \text{mv}_\text{f}^two-\frac{one}{2} \text{mv}_\text{i}^2[/latex].
The work-energy theorem can be derived from Newton'south 2nd constabulary.
Piece of work transfers energy from one place to another or 1 grade to some other. In more than general systems than the particle arrangement mentioned here, work tin can change the potential free energy of a mechanical device, the heat energy in a thermal system, or the electrical energy in an electrical device.

Key Terms

torque: A rotational or twisting effect of a force; (SI unit newton-meter or Nm; regal unit of measurement human foot-pound or ft-lb)

The Piece of work-Energy Theorem

The principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the modify in the kinetic energy of the particle. This definition can be extended to rigid bodies by defining the work of the torque and rotational kinetic energy.

Kinetic Energy: A forcefulness does piece of work on the cake. The kinetic free energy of the block increases as a result by the amount of piece of work. This relationship is generalized in the work-energy theorem.

The work W done by the net force on a particle equals the change in the particle's kinetic free energy ThouE:

[latex]\text{W}=\Delta \text{KE}=\frac{1}{2} \text{mv}_\text{f}^ii-\frac{1}{2} \text{mv}_\text{i}^ii[/latex]

where v_i and v_f are the speeds of the particle before and afterwards the awarding of strength, and m is the particle's mass.

Derivation

For the sake of simplicity, we volition consider the case in which the resultant force F is constant in both magnitude and direction and is parallel to the velocity of the particle. The particle is moving with abiding dispatch a along a direct line. The relationship between the net force and the dispatch is given past the equation F = ma (Newton's 2nd police), and the particle's displacement d, tin can be determined from the equation:

[latex]\text{v}_\text{f}^ii = \text{v}_\text{i}^ii + 2\text{ad}[/latex]

obtaining,

[latex]\text{d}=\frac{\text{v}_\text{f}^2-\text{v}_\text{i}^2}{2\text{a}}[/latex]

The work of the net force is calculated as the product of its magnitude (F=ma) and the particle'south displacement. Substituting the above equations yields:

[latex]\text{Westward}=\text{Fd}=\text{ma}\frac{\text{v}_\text{f}^2-\text{v}_\text{i}^2}{2\text{a}}=\frac{1}{2} \text{mv}_\text{f}^ii-\frac{1}{2} \text{mv}_\text{i}^ii=\text{KE}_\text{f}-\text{KE}_\text{i}=\Delta \text{KE}[/latex]