#### Transcript Energy - Edublogs

Work, Energy and Power AP style Energy Energy: the currency of the universe. Everything has to be “paid for” with energy. Energy can’t be created or destroyed, but it can be transformed from one kind to another and it can be transferred from one object to another. How do you know an object has mechanical energy (kinetic / potential)? If it can change itself or change its environment, then it certainly has energy. • Doing WORK is one way to transfer energy from one object to another. Work = Force x displacement W = Fd • Unit for work is Newton x meter. One Newton-meter is also called a Joule, J. Work = Force x displacement • Work is not done unless there is a displacement. • If you hold an object a long time, you may get tired, but NO work was done. • If you push against a solid wall for hours, there is still NO work done. • For work to be done, the displacement of the object must be in the same direction, as the applied force. They must be parallel. • If the force and the displacement are perpendicular to each other, NO work is done by the force. So, using vector multiplication, W=F•d (In many university texts, as well as the AP test, the displacement is represented by “s” and not “d” W=F•s An object is subject to a force given by F = 6i – 8j as it moves from the position r = -4i + 3j to the position r = i + 7j What work was done by this force? First find the displacement, s s = Dr = rf – ro = (i + 7j) - (-4i + 3j) = 5i + 4j Then, do the dot product W = F · s (6i – 8j) · (5i + 4j) = 30 – 32 = -2J of work • In lifting a book, the force exerted by your hands is upward and the displacement is upward- work is done. • Similarly, in lowering a book, the force exerted by your hands is still upward, and the displacement is downward. • The force and the displacement are STILL parallel, so work is still done. • But since they are in opposite directions, now it is NEGATIVE work. • On the other hand, while carrying a book down the hallway, the force from your hands is vertical, and the displacement of the book is horizontal. • Therefore, NO work is done by your hands. Your Force Vertical component of d • So,….while climbing stairs or walking up an incline, only the vertical component of the displacement is used to calculate the work done in moving the object from the bottom to the top. Horizontal component of d Example How much work is done to carry a 5 kg cat to the top of a ramp that is 7 meters long and 3 meters tall? W = Force x displacement Force = weight of the cat d = height NOT length W = mg x h W = 5 x 10 x 3 W = 150 J 3m Example Displacement = 20 m A boy pushes a lawnmower 20 meters across the yard. If he pushed with a force of 200 N and the angle between the handle and the ground was 50 degrees, how much work did he do? F cos q q F W = (F cos q )d W = (200 cos 50) 20 W = 2571 J NOTE: If while pushing an object, it is moving at a constant velocity, the NET force must be zero. So….. Your applied force must be exactly equal to any resistant forces like friction. How much work do you do to carry a 30 kg cat from one side of the room to the other if the room is 10 meters long? ZERO, because your Force is vertical, but the displacement is horizontal. • Energy and Work have no direction associated with them and are therefore scalar quantities, not vectors. Power The rate at which work is done. 1. Power = Work ÷ time Unit for power = J ÷ s = Watt, W What is a Watt in “fundamental units”? Since work is also the energy transferred or transformed, “power” is the rate at which energy is transferred or transformed. 2. P = Energy ÷ time This energy could be in ANY form: heat, light, potential, chemical, nuclear Since NET work = D K, 3. P = DK ÷ t And yet another approach: P = W ÷ t = (F·d) ÷ t = F · (d ÷ t) P=F·v Kinetic Energy the energy of motion K = ½ mv2 Where does K = ½ mv2 come from?? Did your amazing teacher just arbitrarily make that equation up?? Hmmm… The “Work- Kinetic Energy Theorem” Worknet = DK Fnet •d = DK = ½ mvf2 - ½ mvo2 You are supposed to be able to use Fnet = ma, kinematics equations and Wnet = Fnet •d to derive the work-kinetic energy theorem….. So…. do it!! Then you’ll see where the equation we call “kinetic energy” comes from. (Hint: start with Fnet = ma and use the kinematics equation that doesn’t involve time.) Potential Energy Stored energy It is called potential energy because it has the potential to do work. • Example 1: Spring potential energy in the stretched string of a bow or spring or rubber band. SPE = ½ kx2 • Example 2: Chemical potential energy in fuels- gasoline, propane, batteries, food! • Example 3: Gravitational potential energy- stored in an object due to its position from a chosen reference point. Gravitational potential energy GPE = weight x height GPE = mgh • The GPE may be negative. For example, if your reference point is the top of a cliff and the object is at its base, its “height” would be negative, so mgh would also be negative. • The GPE only depends on the weight and the height, not on the path that it took to get to that height. Work and Energy Often, some force must do work to give an object potential or kinetic energy. You push a wagon and it starts moving. You do work to stretch a spring and you transform your work energy into spring potential energy. Or, you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy. Work = Force x distance = change in energy Example How much more distance is required to stop if a car is going twice as fast? Fd = ½ mv2 The work done by the brakes = the change in the kinetic energy With TWICE the speed, the car has FOUR times the kinetic energy. Therefore it takes FOUR times the stopping distance. The Work-Kinetic Energy Theorem NET Work done = D Kinetic Energy Wnet = ½ mv2f – ½ mv2o Varying Forces The rule is… “If the Force varies, you must integrate!” If the force varies with displacement , in other words Force is a function of displacement, you must integrate to find the work done. If the force is a function of velocity, you must integrate to find the power output. Examples: If F(x) = 5x3, what work is done by the force as the object moves from x = 2 to x = 5? If F(v) = 4v2, what power was developed as the velocity changed from 3 m/s to 7 m/s? Examples of Integration An object of mass m is subject to a force given by F(x) = 3x3 N. What is the work done by the force? 3 4 W F x dx 3 x dx x 4 3 Examples of Integration At object, moving along the x-axis, has a velocity of 3m/s when it passes the origin at t = 0s. It has an acceleration given by a(t) = 1.5t m/s2. Find v(t) and x(t) 15 . 2 v (t ) a(t )dt 15 . tdt t C 0.75t 2 3 2 0.75 2 x(t ) v (t )dt (0.75t 3)dt t 3t C 0.375t 2 3t 0 2 Examples of Integration A particle of mass m moves along the y-axis as y(t)=at4-bt2+2t, where a and b are constants, due to a constant applied force, Fy(t) What is the power, P(t) delivered by the force? P = F·v We need to find both v and F! v t 4at 3 2bt 2 a(t )12at 2 2b F ma m(12at 2 2b) P Fy v y m(12at 2b)(4at 2bt 2) 2 3 Examples of Definite Integration An object is moving along the x-axis with a velocity given by v(t)=-2t2+8t-6 where t is in seconds. What was its displacement from t = 2s to t = 6s? 6 6 2 8 x (t ) v (t )dt ( 2t 2 8t 6)dt t 3 t 2 6t 3 2 2 2 6 2 2 8 2 8 ( 6 3 6 2 66)( 2 3 2 2 62) 45.33m 3 2 3 2 Examples of Definite Integration To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 – x. What work is done to stretch the spring 2 meters beyond its equilibrium position? 2m W 0 2m 4 3 1 2 F ds (4 x x )ds x x 3 2 0 2 2m 0 4 3 1 2 4 2 1 ( 2 2 )( 0 0)8.667J 3 2 3 2 Examples of Definite Integration A 2.88kg particle starts from rest at x = 0 and moves under the influence of a single force, Fx= 6 + 4x -3x2 where Fx is in Newtons and x is in meters. Find the power delivered to the particle when it is at x = 3m. P = F·v, but what is the velocity at x = 3m? Hmmm…. W F( x )dx W DK so..... and 3 xf 1 mv 2 1 mv 2 F ( x ) dx D K f o 2 2 (6 4 x 3 x 0 xo 3 (64x 3x 2 )dx 6x 0 4 2 3 3 x x 2 3 3 0 2 )dx DK 12 mv x2 3 4 3 (63 3 2 3 3 )0 9J 12 mv x2 3 2 3 v x 3 2.5m / s and Fx 3 64333 2 9N so P F v 92.5 22.5W