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WORK, ENERGY & MOMENTUM. WORK & KINETIC ENERGY Work, W: using a force, F, to displace an object a distance, d unit: Joule (1 J = 1 Nm) W = Fd W = (Fcos.

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Presentation on theme: "WORK, ENERGY & MOMENTUM. WORK & KINETIC ENERGY Work, W: using a force, F, to displace an object a distance, d unit: Joule (1 J = 1 Nm) W = Fd W = (Fcos."— Presentation transcript:

1 WORK, ENERGY & MOMENTUM

2 WORK & KINETIC ENERGY Work, W: using a force, F, to displace an object a distance, d unit: Joule (1 J = 1 Nm) W = Fd W = (Fcos  )d W = 0

3 WORK & KINETIC ENERGY Work done by any force: W = Fdcos  can be positive, negative, or zero Ex: sled sliding down a hill gravity does positive work friction does negative work normal force does no work d  d F

4 WORK & KINETIC ENERGY Power, P: the time rate at which work is done P = W/t unit: Watt, W (1 W = 1 J/s) (1 J/s = 1 Nm/s) english unit: horsepower, hp (1.00 hp = 746 W) P = Fv Lift Weight Thrust Drag

5 WORK & KINETIC ENERGY Kinetic Energy, K: energy of motion Energy: the ability to do work K = ½mv 2 unit: Joule scalar quantity – amount only – direction doesn’t matter can only be zero or positive – never negative

6 WORK & KINETIC ENERGY

7 Work/Energy Theorem: net work done on an object is equal to the total change in kinetic energy of the object W net = K f – K i F net dcos  = ½mv f 2 – ½mv i 2

8 WORK & KINETIC ENERGY Net work determines the change in an object’s motion positive work = increase in kinetic energy (speed up) Ex: throwing a ball negative work = decrease in kinetic energy (slow down) Ex: catching a ball zero work = no change in kinetic energy Ex: weightlifting

9 PHYSICS UNIT 4: ENERGY & MOMENTUM

10 POTENTIAL ENERGY & CONSERVATION Potential Energy, U: energy of position Gravitational PE: energy of position due to gravity force U g = mgh h: height, measured from origin (reference point) unit: Joule, J Scalar Quantity - can be positive, zero, or negative depending on choice of origin

11 POTENTIAL ENERGY & CONSERVATION pendulu m: U   K K   U the amount stays the same

12 POTENTIAL ENERGY & CONSERVATION Conservation of Mechanical Energy: a system's total mechanical energy (K+U) stays constant if there is no friction K i + U i = K f + U f However, if there is friction, some K will be turned into other energy forms - heat, sound, etc. K i + U i = K f + U f + W lost mechanical energy is not conserved total energy is still conserved

13 Example: a Mass in Free Fall K i + U i = K f + U f ½mv i 2 + mgh i = ½mv f 2 + mgh f Cons. Of Energy

14 POTENTIAL ENERGY & CONSERVATION Example: a Mass on a Horizontal Spring ½mv i 2 + ½kx i 2 = ½mv f 2 + ½kx f 2 K i + U i = K f + U f

15 POTENTIAL ENERGY & CONSERVATION

16 PHYSICS UNIT 4: ENERGY & MOMENTUM

17 QUIZ 4.1 Joe throws a ball straight up into the air, and catches it on the way back down. (a) Draw a graph showing the kinetic energy of the ball throughout its flight. (b) Draw a graph showing the gravitational potential energy of the ball throughout its flight. (c) Draw a graph showing the total energy of the ball throughout its flight.

18 PHYSICS UNIT 4: ENERGY & MOMENTUM

19 QUIZ 4.2 (a) Tell what kinds of energy a pole vaulter has at each of the four points labeled on the picture above (point 4 is just before hitting the mat) (b) After the pole vaulter hits the mat, his total energy is zero. Where did all his energy go?

20 PHYSICS UNIT 4: ENERGY & MOMENTUM

21 QUIZ 4.3 A roller coaster car, mass 500 kg, starts from rest at the top of a hill 30 m above ground level. Ignore friction. (a) What is the car’s potential energy at the top of the hill? (b) What is the car’s kinetic energy at the bottom of the hill? (c) How fast will the car be going at the bottom of the hill? (d) What is the car’s kinetic energy at the top of the next hill, 10 m above ground level? 147,000 J 24.2 m/s 98,000 J

22 PHYSICS MOMENTUM

23 MOMENTUM & IMPULSE Momentum, p: amount of “umph" an object has (Inertia in Motion) = mv unit p : kg m/s vector quantity - includes direction +2 kgm/s –2 kgm/s

24 MOMENTUM & IMPULSE Impulse, J: A force that acts over a duration of time. J = Ft unit: kg m/s or N s

25 MOMENTUM & IMPULSE Impulses cause a change in momentum. This is known as the Impulse-Momentum Theorem. It is analogous to the Work-Energy Theorem. FΔT = Δp = p f – p i = mv f -mv i unit: kg m/s or N s force of impact, F = -p i /t to decrease force of impact, decrease p i (decrease v before impact) or increase t (catching an egg; stunt falling; air bags)

26 Practice A 2000 kg car going 30 m/s hits a brick wall and comes to rest. (a) What is the car’s initial momentum? (b) What is the car’s final momentum? (c) What impulse does the wall give to the car? (d) If the impact takes 0.5 seconds, what force is exerted on the car? 60,000 kg m/s -120,000 N -60,000 kg m/s 0 kg m/s

27 MOMENTUM & IMPULSE Bouncing vs. Sticking in an impact ex: a 1000 kg car going +10 m/s hits a wall J = p f -p i sticking: p i = +10,000 kgm/s, p f = 0 J = –10,000 kgm/s bouncing: p i = +10,000 kgm/s, p f = – 10,000 kgm/s J = –20,000 kgm/s bouncing off at impact has up to twice the force of sticking

28 MOMENTUM & IMPULSE Law of Conservation of Momentum: total momentum of a system of objects is constant if no outside forces act m i v i = m f v f  if mass increases, velocity decreases (and vice versa)

29 COLLISIONS inelastic collision: objects collide and stick (or collide and deform) momentum is conserved, kinetic energy is not BEFORE = AFTER m 1 v 1 + m 2 v 2 = Mv f (M = m 1 + m 2 ) be sure to include + or – for velocity’s direction

30 COLLISIONS propulsion or explosion: total initial momentum is zero; separated pieces receive equal & opposite momentums, so total final momentum is zero 0 = m 1 v 1f + m 2 v 2f or m 1 v 1f = –m 2 v 2f ex: rocket propulsion, gun recoil

31 COLLISIONS Ex: A 4 kg rifle fires a 0.050 kg bullet, giving the bullet a final velocity of 300 m/s east. What is the recoil velocity of the rifle?

32 COLLISIONS elastic collision: objects collide and bounce off with no loss of energy both momentum and kinetic energy are conserved BEFORE = AFTER m 1 v 1o + m 2 v 2o = m 1 v 1f + m 2 v 2f ½m 1 v 1o 2 + ½m 2 v 2o 2 = ½m 1 v 1f 2 + ½m 2 v 2f 2

33 Useful Equations p = mv J = p f – p i = Ft m 1 v 3 = –m 2 v 4 m 1 v 1 + m 2 v 2 = Mv 3 m 1 v 1 + m 2 v 2 = m 1 v 3 + m 2 v 4


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