Energy is the capacity to do work

Energy is another fundamental concept in the study of physics.  There are many different times of energy, some of which include potential, kinetic, gravitational, electric, etc.  The law of conservation of energy states that the total energy in an isolated system remains constant; no energy can be gained or lost. The equation that expresses the conservation of energy is as follows:

W+ + KEi + GPEi = W- + KEf + GPEf

Kinetic energy and potential energy are two important forms of energy.  Kinetic energy is the energy an object has because of its motion, and can be quantified by the equation KE = mv2/2.  Potential energy is the energy that an object has because of its position, and gravitational potential energy can be quantified by the equation GPE = mgh.  W+ and W- represent work added or removed from a system.  Using these quantities and the formula above, we can solve many different types of problems.

Work is force applied over a distance, and is done using energy.  The formula for work is w = fdcosθ, where f is the force is Newtons, d is the distance in meters, and theta is the angle at which the force is applied.  The work-energy theorem states that Wnet = KEf - KEi, or that the net work is equal to the change in kinetic energy.

Power is the rate at which work is done.  The formula for power is P = w/t = fdcostheta/t = fv, where f refers to force, w refers to work, t refers to time, d refers to distance, theta refers to the angle at which the force is applied, and v refers to the velocity of the object.  Power is measured in Watts, and 746 Watts are equal to 1 horsepower.

Example Problem #1

A ball (2kg) is dropped from a height of 100m.  When the ball is 25m above the ground (h=0m), what is the ball’s speed?

Answer: 38.34 m/s

Explanation: We will use the law of conservation of energy to solve this problem. Since no work is added or lost from the system, we can disregard those two variables in the equation.  Initially, there is no kinetic energy, so we can disregard that variable on the left side of the equation.  We now have the equation:

GPEi = KEf + GPEf

mghi = .5mv^2 + mghf → 2(9.8)(100) = v^2 + 2(9.8)(25)

Solving for v, we get that the velocity of the ball when it is 25m above the ground is 38.34 m/s.