Work and Kinetic Energy §

Introduction §

Work and Kinetic Energy are fundamental concepts in classical mechanics, describing how forces cause motion and changes in the energy of a system. This note will explore these concepts, their mathematical formulations, and their interrelationship.

Work §

Work is defined as the process of energy transfer to or from an object via the application of force along a displacement. In mathematical terms, work ($W$) done by a constant force ($\mathbf{F}$) on an object during its displacement ($\mathbf{d}$) is given by: $W=\mathbf{F}\cdot \mathbf{d}=Fd\cos (\theta )$ where $\theta$ is the angle between the force vector and the displacement vector.

Types of Work §

1. Positive Work: When force and displacement are in the same direction.
2. Negative Work: When force and displacement are in opposite directions.
3. Zero Work: When force is perpendicular to displacement, or when there is no displacement.

Kinetic Energy §

Kinetic energy ($KE$) is the energy that an object possesses due to its motion. It is defined as: $KE=\frac{1}{2}m{v}^2$ where $m$ is the mass of the object and $v$ is its velocity.

Work-Energy Theorem §

This theorem states that the work done by all forces acting on an object is equal to the change in its kinetic energy. Mathematically, it is expressed as: $W=\Delta KE=K{E}_{final}-K{E}_{initial}$

Applications and Examples §

1. Rolling Ball: A ball rolling down an incline gains kinetic energy as work is done by gravity.
2. Bow and Arrow: When a bow is drawn, work is done to store energy in the bow, which is then transferred to the arrow as kinetic energy upon release.

Test Questions §

1. [Basic] Question: If a force of 10 N is applied to a box, moving it 5 m in the same direction, what is the work done on the box? Back: $W=Fd=10\times 5=50$ Joules.
2. [Basic] Question: A car of mass 1000 kg increases its speed from 10 m/s to 20 m/s. Calculate the change in its kinetic energy. Back: $\Delta KE=\frac{1}{2}m(v_{final}^2-v_{initial}^2)=\frac{1}{2}\times 1000\times (20^2-10^2)=150,000$ Joules.
3. [Basic] Question: How much work is done by a force that displaces an object by 4 meters at an angle of 60 degrees to the direction of force, if the force magnitude is 20 N? Back: $W=Fd\cos (\theta )=20\times 4\times \cos (60^{\circ })=40$ Joules.