Work, Energy & Power. AP Physics B

Transcription

1 ork, Energy & Power AP Physics B

2 There are many dierent TYPES o Energy. Energy is expressed in JOULES (J) 4.19 J = 1 calorie Energy can be expressed more speciically by using the term ORK() ork = The Scalar Dot Product between Force and Displacement. So that means i you apply a orce on an object and it covers a displacement you have supplied ENERGY or done ORK on that object.

3 Scalar Dot Product? A product is obviously a result o multiplying 2 numbers. A scalar is a quantity with NO DIRECTION. So basically ork is ound by multiplying the Force times the displacement and result is ENERGY, which has no direction associated with it. = r F r x rr Fx cosθ A dot product is basically a CONSTRAINT on the ormula. In this case it means that F and x MUST be parallel. To ensure that they are parallel we add the cosine on the end. = Fx Area = Base x Height

4 ork The VERTICAL component o the orce DOES NOT cause the block to move the right. The energy imparted to the box is evident by its motion to the right. Thereore ONLY the HORIZONTAL COMPONENT o the orce actually creates energy or ORK. hen the FORCE and DISPLACEMENT are in the SAME DIRECTION you get a POSITIVE ORK VALUE. The ANGLE between the orce and displacement is ZERO degrees. hat happens when you put this in or the COSINE? hen the FORCE and DISPLACEMENT are in the OPPOSITE direction, yet still on the same axis, you get a NEGATIVE ORK VALUE. This negative doesn't mean the direction!!!! IT simply means that the orce and displacement oppose each other. The ANGLE between the orce and displacement in this case is 180 degrees. hat happens when you put this in or the COSINE? hen the FORCE and DISPLACEMENT are PERPENDICULAR, you get NO ORK!!! The ANGLE between the orce and displacement in this case is 90 degrees. hat happens when you put this in or the COSINE?

5 The ork Energy Theorem Up to this point we have learned Kinematics and Newton's Laws. Let 's see what happens when we apply BOTH to our new ormula or ORK! 1. e will start by applying Newton's second law! 2. Using Kinematic #3! 3. An interesting term appears called KINETIC ENERGY or the ENERGY OF MOTION!

6 The ork Energy Theorem And so what we really have is called the ORK-ENERGY THEOREM. It basically means that i we impart work to an object it will undergo a CHANGE in speed and thus a change in KINETIC ENERGY. Since both ORK and KINETIC ENERGY are expressed in JOULES, they are EQUIVALENT TERMS! " The net ORK done on an object is equal to the change in kinetic energy o the object."

7 Example =Fxcosθ A 70 kg base-runner begins to slide into second base when moving at a speed o 4.0 m/s. The coeicient o kinetic riction between his clothes and the earth is He slides so that his speed is zero just as he reaches the base (a) How much energy is lost due to riction acting on the runner? (b) How ar does he slide? a) = 0 = = K J mv 2 o x = = (70)(4) F = 1.17 m 2 x cosθ F = µ F mg n µ = (0.70)(70)(9.8) = N (480.2) x(cos180)

8 Example A 5.00 g bullet moving at 600 m/s penetrates a tree trank to a depth o 4.00 cm. (a) Use the work-energy theorem, to determine the average rictional orce that stops the bullet.(b) Assuming that the rictional orce is constant, determine how much time elapses between the moment the bullet enters the tree and the moment it stops moving = K = 0 1 (0.005)(600) 2 = -900 J riction 2 F = F = ma 22,500 = (0.005) a a = NET 4.5x10 6 m/s/s ( ) 900 = F 0.04 F = F x cosθ = 22,500 N 6 v = vo + at 0 = ( 4.5x10 ) t t = 1.33x10-4 s

9 Liting mass at a constant speed Suppose you lit a mass upward at a constant speed, v = 0 & K=0. hat does the work equal now? Since you are liting at a constant speed, your APPLIED FORCE equals the EIGHT o the object you are liting. Since you are liting you are raising the object a certain y displacement or height above the ground. hen you lit an object above the ground it is said to have POTENTIAL ENERGY

10 Suppose you throw a ball upward hat does work while it is lying through the air? GRAVITY Is the CHANGE in kinetic energy POSITIVE or NEGATIVE? NEGATIVE Is the CHANGE in potential energy POSITIVE or NEGATIVE? = K = U K ( K K ) = U U o = U o o o K + K = U U U + K = U + K Energy BEFORE = o o Energy AFTER POSITIVE

11 ENERGY IS CONSERVED The law o conservation o mechanical energy states: Energy cannot be created or destroyed, only transormed! Energy Beore Am I moving? I yes, K o Am I above the ground? I yes, U o Energy Ater Am I moving? I yes, K Am I above the ground? I yes, U

12 Energy consistently changes orms

13 Energy consistently changes orms Am I above the ground? NO, h = 0, U = 0 J Am I moving? Yes, v = 8 m/s, m = 60 kg K K = 1 mv 1 (60)(8) 2 2 = 1920J 2 2 Position m v U K ME (= U+K) 1 60 kg 8 m/s 0 J 1920 J 1920 J

14 Energy consistently changes orms Energy Beore K O = Energy Ater = U + K 1920= (60)(9.8)(1) + (.5)(60)v = v = 30v = v 2 v = 6.66 m/s Position 1 2 m 60 kg 60 kg v U K ME 8 m/s 0 J 1920 J 1920 J 6.66 m/s 588 J 1332 J 1920 J

15 Energy consistently changes orms Am I moving at the top? No, v = 0 m/s E B = E A Using position 1 K o = U 1920 = mgh 1920 =(60)(9.8)h h = 3.27 m Position m v U K ME 1 60 kg 8 m/s 0 J 1920 J 1920 J 2 60 kg 6.66 m/s 588 J 1332 J 1920 J 3 60 kg 0 m/s 1920 J 0 J 1920 J

16 Example A 2.0 m pendulum is released rom rest when the support string is at an angle o 25 degrees with the vertical. hat is the speed o the bob at the bottom o the string? θ Lcosθ h L h = L Lcosθ h = 2-2cosθ h = m E B = E A U O = K mgh o = 1/2mv 2 gh o = 1/2v = v m/s = v

17 Power One useul application o Energy is to determine the RATE at which we store or use it. e call this application POER! As we use this new application, we have to keep in mind all the dierent kinds o substitutions we can make. Unit = ATT or Horsepower