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ENERGY AND WORK
In metric SI units, the potential energy EPF of a body of mass M kilograms elevated to a height of S meters, is MgS joules. After it has fallen a distance S, the kinetic energy gained will thus be MgS joules. Another type of potential energy is elastic potential energy, such as possessed by a spring that has been compressed or extended. The amount of work in ft lbs done in compressing the spring S feet is equal to KS2/2, where K is the spring constant in pounds per foot. Thus, when the spring is released to act against some resistance, it can perform KS2/2 ft-lbs of work which is the amount of elastic potential energy EPE stored in the spring. Using metric SI units, the amount of work done in compressing the spring a distance S meters is KS2/2 joules, where K is the spring constant in newtons per meter. Work Performed by Forces and Couples.—The work U done by a force F in moving an object along some path is the product of the distance S the body is moved and the component F cos α of the force F in the direction of S. U = FS cos α where U = work in ft-lbs; S = distance moved in feet; F = force in lbs; and α = angle between line of action of force and the path of S. If the force is in the same direction as the motion, then cos α = cos 0 = 1 and this formula reduces to: U = FS Similarly, the work done by a couple T turning an object through an angle θ is: U = Tθ where T = torque of couple in pounds-feet and θ = the angular rotation in radians. The above formulas can be used with metric SI units: U is in joules; S is in meters; F is in newtons, and T is in newton-meters. Relation between Work and Energy.—Theoretically, when work is performed on a body and there are no energy losses (such as due to friction, air resistance, etc.), the energy acquired by the body is equal to the work performed on the body; this energy may be either potential, kinetic, or a combination of both. In actual situations, however, there may be energy losses that must be taken into account. Thus, the relation between work done on a body, energy losses, and the energy acquired by the body can be stated as: Work Performed – Losses = Energy Acquired U – Losses = E T Example 1:A 12-inch cube of steel weighing 490 pounds is being moved on a horizontal conveyor belt at a speed of 6 miles per hour (8.8 feet per second). What is the kinetic energy of the cube? Since the block is not rotating, Formula (a) for the kinetic energy of a body moving with pure translation applies: WV 2 Kinetic Energy = ----------2g 490 × ( 8.8 ) 2 = ------------------------------ = 590 ft-lbs 2 × 32.16 A similar example using metric SI units is as follows: If a cube of mass 200 kg is being moved on a conveyor belt at a speed of 3 meters per second, what is the kinetic energy of the cube? It is: Kinetic Energy = 1⁄2 MV 2 = 1⁄2 × 200 × 3 2 = 900 joules