Let's plot F s vs X. In terms ofx, F S = KX. Let's assume the spring is stretched X meters as shown

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1 Elastic potential energy Let's plot F s vs X F s The work done in stretching the spring can be calculated using the area under a force vs distance curve. Therefore the spring has stored elastic potential energy equal to the area under this curve. Elastic potential energy, E e = the area under the curve. The area is a triangle. A= b h/2 Therefore E e = (X ) ( KX ) / 2 E e = KX 2 / 2 Let's assume the spring is stretched X meters as shown In terms ofx, F S = KX The elastic potential energy formula is... E p = KX 2 / 2 c Try this! The spring in the previous example has a force constant of 62.5 N/m and is stretched 8.00 cm. How much elastic potential energy does this stretched spring have?. Given: K = 62.5 N/m X = 8.00 cm or m. Unknown: E e =?. Formula: E e = KX 2 / 2. Sub: E e =(62.5)(0.0800) 2 /2 = J» Answer: The spring has J of elastic potential energy. It is able to do J of work.

2 Work Work. So Work can be defined mathematically as... W = Ad F or Ad X F cos9. SI unit is the Joule = N m = ( kg m /s 2 ) m = kg m 2 / s 2. Note that "F" in the formula can be any kind of force. Both the force doing the work and the object that the work is being done on must be specified or understood. Work is the product of the magnitude of the displacement moved times the component of a force in the direction of the displacement. Defining equation: W = Ad X F cosg where e is the smallest angle between Ad and F in a " tail-to-tail" diagram d a) What is the box by the applied force? W = Ad X F cos0 = 5.00 m X 100.ON COS = 400 J a. 0 = 36.9 W = Ad X F coso Work W can be written using a special mathematical "vector operator" W = Ad F So Ad F = I Ad I X I F I cosg note that stands for "dot" or "scalar" product Since Work is a product of magnitudes only, it is a scalar quantity.

3 c) What is the box by the kinetic friction force? f K = 22.0 N el F a =100N b) What is the box by the net force? W = Ad X F cos9 =5.00 m X 22 N cos 180 C = -110J Note that f k has 6 = 180, so a short-cut formula can be used W = - f,, d f K = 22.0 N 9 = 180 The negative sign means energy is lost to the surroundings as heat. F net =100cos = 58.0 N [E] a d) What is the box by the weight? W = Ad X F cos9 = 5.00 m X 480N cos 90 = OJ Note that if 6 = 90, no work is done f K = 22.0 N F 0 =480 N y F a =100N b) What is the box by the net force? W = Ad X F cose = 5.00 m X 58.0 N cos O c = 290 J Note that if 6 = 0, then a short-cut formula can be used W = F d F ne t = 58.0 N The work done by the net force is sometimes called the "net work done" or W net

4 Energy of Motion Conditions where no work is done Moving objects can collide with other bodies and exert a "force through a distance" on these bodies, so moving objects are able to do work on other bodies. Energy that an object possesses by virtue of its motion is called kinetic energy. The symbol for kinetic energy is E k. E k =mv 2 /2 1. if e = If F = 0 N 3. If I Ad I = 0 m Try this example: Compare the kinetic energy of a 5055 kg truck and a 20 g bee moving at 36.0 km/h. E k = mv 2 /2 m = 5055 kg Truck v = 36.0 km/h = 10.0 m/s m = kg Bee v = 36.0 km/h = 10.0 m/s E k = mv 2 /2 = 5055(10.0) 2 /2 E k = mv 2 /2 = 0.020(10.0) 2 /2 = 2.5X10 5 J =1.0J Energy Energy is the ability to do work Since energy is defined in terms of "work", it is a scalar quantity. Like work, the SI unit for any type of energy is the Joule.

5 More About the formula for gravitational potential energy E g = mgh Note that the height or "h" and therefore E g depends on the reference level used to specify E g = 0 or the zero level of gravitational potential energy. Usually "ground level" is the specified zero reference level but not always. The zero reference level could be the top of a table or the bottom of the swing of a pendulum. Note "h" can be + or - depending on whether it is "above 1 or "below" the zero reference level. Try this example #2 : An 8.00 kg bird has a kinetic energy of 576 J. How fast is it moving?. Given: m = 8.00 kg E k = 576 J. Unknown: v =?. Formula: E k =mv 2 /2 so v = (2 E k /m ) 1/2. Sub: v = (2X576/8.00) 1/2 = 12m/s Try this example: A 14 kg boulder is on a ledge 6.0 m above ground level and 4.0 m below the top of a cliff as shown. Find the gravitational potential energy of the boulder relative to the... a) ground b) top of the cliff c) ledge Cliff top a) E = mgh = 14X10.0X6.0 = 840 J b) E = mgh = 14X10.0 (-4) = -560J c) E = mgh = 14X10.0(0) = OJ = 14kg Temperature is another scalar quantity like E g and it can have negative values too. It just depends on the zero reference level as well. The freezing point of water is 0 C in the Celsius scale but 273 K in the Kelvin scale. ground Energy of raised position An object that is raised above a certain reference level "is able" or has "potential" to do work. If the object falls, the force of earth's gravity can exert a "force through a distance" on that object and do work on it. The energy of raised position is called gravitational potential energy. Its symbol is E y The formula for gravitational potential energy is E g - mgh

6 F s a X F s =KX More on Hooke's Law K, the constant of proportionality, is called the force constant of the spring or Hooke's constant of the spring or just the spring constant. K depends on the "stiffness" of the spring involved If F s = KX, then K = F s / X and SI unit for K is N/m. Energy stored in a stretched or compressed spring Like a slingshot, a stretched or compressed spring can exert a "force through a distance" and "is able to do work" on an object. Therefore, a deformed spring has energy. The energy of a deformed spring is called elastic potential energy. The symbol for elastic potential energy is E e. To understand further, we need to know about forces on deformed springs. Hooke's Law Example: A kg mass is placed on a vertical spring at equilibrium. The mass stretches the spring 8.00 cm as shown and then stays at rest. Find the force constant of the spring. F nety = F S + F =0 +KX = 0 +K(.0800) = 0 K(.0800) = 5.00 K= 5.00 /.0800 K= 62.5 N/m This k value tells us that this spring requires 62.5 N of force to stretch or compress it by 1 meter. However, most strings have a threshold limit of deformation, beyond which Hooke's law does not hold. rest X = 8.00 cm 1 F > sf m = ffi :0.500 kg Forces needed to "deform" a spring X = amount of compression or stretch of a spring from its rest position (deformation) F s = magnitude of force needed to deform a spring Robert Hooke discovered that F s a X Hooke's law Equilibrium or rest position compression