Underlying all of physics, at the very bedrock of the foundation, are several "; conservation laws";. We have seen one, the conservation of momentum, in the last chapter. We saw how in any interaction between masses, without exception, a quantity called momentum remained unchanged.

Through chemistry you have probably learned of the principle of conservation of mass. That is, matter is neither created nor destroyed, but must remain a constant in any interaction. In this chapter we consider the conservation of energy. It is not as neatly contained as the one for momentum, because energy can exist in so many different forms: light, heat, electricity, gravity, and motion to name a few. Initially we will consider only a few of these, but they will enable us to analyze a large number of physical phenomena.

The first form of energy is work. Work is defined simply as a force times the distance over which that force acts.

W = F * s

Thus if a 4 pound book is lifted 3 feet, the work done is 12 foot-pounds. If it takes 7 newtons to slide a box 5 meters along a floor, the work done is 35 newton-meters.

It's important to note that if there is no motion there is no work. This conflicts with our everyday English use of the word work which is very personally related. Yowork if your muscles get tired, or if you start to sweat, or if you study late into the night, or if you try hard at a crossword puzzle. In physics these factors mean nothing. A force must be exerted, and there must be a component of motion parallel to the force to have work done. (Thus an ice cube sliding on glass does so without any work being done. While the force of gravity on the cube is down, its motion is sideways. There is no motion parallel to the force and therefore no work done.)

1. A 5.6 newton weight is lifted 2.2 meters. How much work is done?

W = F * s (5.6nt)(2.2m) W = 12.3 nt - m

note: A nt-m is such a common and well-beloved unit, it has been given its own name, the ";joule";, named after James Prescott Joule.

2. A 1.7 slug mass is lifted 4.0 feet up. How much work is done?

The applied force is F= mg = (1.7 slug)(32 ft/sec2)= 54.4 lb W = F * s = (54.4 lb)(4 ft) W = 218 ft - lb

3. A 6 kg mass rests on a surface of µ = 0.2. How much work is required to slide the mass 1.4 meters?

Ff = µN = (.2)(6 * 9.8)= 11.76 nt W = F * s = (11.76)(1.4) = 16.5 nt - m W = 16.5 J

4. A 5 pound weight slides 7 feet along a frictionless surface. How much work is done?

W = F * s = (0)(7ft) W = 0 J (No force is required on a frictionless surface)

5. A 0.83 slug mass rests on a surface of µ = 0.3. How much work is required to slide it 4.0 feet?

Ff = µN = (.3)(.83 * 32) W = F * s = (.3)(.83 * 32)(4.0) W = 31.9 ft - lb

6. A Perry N. Thrust pushes with a 35 newton force at 40° on a block as shown. How much work does he do in sliding the block 78 cm?

W = F * s = (35cos40)(.78m) W = 20.9 joules

You're probably wondering...if the English unit of work is the foot-pound, and the SI unit of work is the newton-meter of joule, what is the CGS unit of work? Since work is force times distance in any system, the dyne-centimeter is the work unit, fondly known as the erg. This is an exceedingly small unit of energy.

7. How much energy is required to lift a 2 gram paper clip 5 cm off a table?

W = F * s = (mg)(s) = (2gm * 980 cm/sec2)(5 cm) W = 9.80 * 10 3

8. How far must a 0.86 gm fly be dragged along a table to consume 100 ergs of energy? The table has a coefficient of friction µ = 0.2.

W = F * s = (µg)(s) 100 = (.2 * .86 * 980)(s) s = .593 cm

9. A 3 kg mass rests on a 34° plane with friction µ = 0.2. How much work is required to slide the mass 2.7 m up the incline?

To just barely slide the block, F - Fg - Ff = 0 F = µmgcosØ + mgsinØ F = .2(3 * 9.8)cos34 + (3 * 9.8)sin34 F = 21.3 nt W = F * s W = (21.3 nt)(2.7 m) W = 57.5 J

10. A 17 lb weight rests on the top of a 22° incline with coefficient of friction 0.6. How much work is required to slide it 4.4 ft down the plane?

F + Fg - Ff = 0 F + 17sin22 - .6(17cos22) = 0 F = 3.09lb W = F * s = (3.09)(4.4) W = 13.6 ft-lb

11. A mass m rests on a plane inclined at angle q with coefficient of friction µ. How much work is required to slide it a distance s?

F - Fg - Ff = 0 F - mgsinØ - µmgcosØ = 0 W = F * s W = mgs(µcosØ + sinØ)

12. A mass m is lifted to a height h. How much work is required?

W = F * s W = mg * h W = mgh

From problem 12 we can deduce a new form of energy. In lifting the mass we lose nothing to friction, so all the energy we exert must be stored in some new form. This is called potential energy, or in this case gravitational potential energy. It is simply a means of storing energy by placing objects in high places. (Mother Nature does this all the time by accumulating enormous masses of water in clouds. As the water flows back to the sea, clever humans skim off a little of the energy in the form of hydroelectricity.)

Thus we can say:

PEg = mgh

13. What is the potential energy of a 6 kg mass 4 meters above the ground?

PEg = mgh PEg = 6 * 9.8 * 4 PEg = 235 J

14. What is the potential energy of a 4.0 slug mass 3.5 ft in the air?

PEg = mgh PEg = 4 * 32 * 3.5 PEg = 448

15. Find the potential energy of a 9.2 slug boulder in a 20 foot deep hole.

PEg = mgh PEg = 9.2 * 32 * (-20) PEg = -5.89 * 103ft-lb

16. How much mass must be placed on a 3.7 m high shelf to have 25 joules of energy?

PEg = mgh m = PE / gh m = 25 / 9.8 * 3.7 m = .689 kg

Potential energy may be converted to another kind of energy simply by tipping the mass off the shelf and letting it fall to the ground. Just before it hits it will have no more potential energy, but it will have acquired energy of motion kinetic energy. To find the proper expression for kinetic energy the speed of the falling mass must be found.

In falling from a height h a mass will acquire a speed	v = (2gh)1/2
Rearranging, we get	h = v2/2g
The original potential energy, then, becomes	mgh = mg(v2/2g) = 1/2 mv2
Thus we say that kinetic energy is	KE = 1/2 mv2

17. What are the kinetic energies of
a) a 6 kg whippet running at 4 m/sec?
b) a 1.4 slug carton of Cheeze-Whiz traveling at 3 ft/sec?
c) a 45 gm Twinkie falling at 27 cm/sec?

a) KE = 1/2mv2 = 1/2 (6 kg)(4 m/sec)2 KE = 48 kgm2/sec2 KE = 48.0 J b) 1/2mv2 = 1/2 (1.4 slug)(3 ft/sec)2 KE = 6.30 ft-lb c) 1/2mv2 = 1/2 (45 slug)(27 cm/sec)2 KE = 1.64 * 104

18. What is the kinetic energy of a 135 pound sack of compost catapulted at 19 ft/sec?

KE = 1/2mv2 = 1/2 (135/32) * 192 KE = 761 slug ft2/sec2 KE = 761 ft-lb

19. Find the kinetic energy of a 270 pound wappati running at 11 ft/sec.

KE = 1/2mv2 KE = 1/2 (230/32 slug) * (11 ft/sec)2 KE = 510 ft-lb

20. How fast must a 7,850 nt car be traveling to have 1.5 x 105 joules of energy?

KE = 1/2mv2 KE = 1.5 * 105 = 1/2 (7850/9.8)v2 v = 19.4 m/sec

21. How fast must a 15 nt weight be moving to have 4.8 joules of energy?

KE = 1/2mv2 v = (2KE/m)1/2 v = (2 * 4.8/ 15 / 9.8)1/2 v = 2.50 m/sec

22. What is the total kinetic energy of each system?

a)	b)	c)
In every case KE = KE1 + KE2 KE = KE1 + KE2 = 1/2(3)(12)2 + 1/2(7)(5)2 KE = KE1 + KE2 = 304 J Energy is not a vector quantity, so direction is not important.

You now have all the tools required to start solving some more interesting problems.

23. A 23 nt force pushes a 6 kg block of cheddar cheese along a surface of µ = 0.2. What speed will the block acquire after traveling 3 meters?

We know a certain amount of work is done on the block: W = F * S. This energy goes into overcoming friction and into kinetic energy of the mass Word done - work of friction = kinetic energy F * s - (µmg)l = 1/2mv2 23 * 3 - .2 * 6 * 9.8 * 3 = 1/2 (6)v2 v = 3.35 m/sec We could have also have said F - µmg = ma 23 - .2 * 6 * 9.8 = 6a a = 1.873 m/sec2 v = (2as)1/2 v = (2(1.873)(3))1/2 v = 3.35 m/sec

24. A 3 kg mass slides along a level frictionless surface at 7 m/sec. It hits a 5 m long region with µ = 0.2. What is its speed at the other side?

KE0 - Wf = KEf 1/2mv02 - µmgl = 1/2mvf2 1/2 (3)(7)2 - .2(3 * 9.8 * 5) = 1/2 mvf2 vf = 5.42 m/sec

25. A 7 slug mass slides along a frictionless track at 17 ft/sec. It rises 4 ft and levels off. What is its new velocity?

KE0 - PEg = KEf 1/2mv02 - mgh = 1/2mvf2 (M'S CANCEL) 1/2 (17)2 - (32 * 4) = 1/2 vf2 vf = 5.42 m/sec

26. A force of 33 nt is applied to a 5 kg mass which slides first along a surface of µ = 0.3 as shown. What is its final velocity?

WFORCE - WFRICTION1 - WFRICTION2 = KEFINAL F * s - µ1mgl1 - µ2mgl2 = 1/2mvf2 33 * 7 - .2(5 * 9.8 * 4) - .3(5 * 9.8 * 3) = 1/2 (5)vf2 vf = 7.69 m/sec

I hope you're wondering what's happening to this work being done by friction. Obviously it's being removed from the mechanical energy of the systems. Rub your hands together hard and you will know it goes into heat. Heat is the final resting place of all forms of energy, and it is in this form that energy is radiated back out into space.

27. A 16 pound blob of clay falls from a 7 ft high shelf and splats onto the floor. How much heat is generated?

PEg » KE » HEAT mgh "arrow 1/2mv2 » HEAT 16 * 7 = 112 ft-lb = HEAT GENERATED

28. A 3.6 slug mass slides with an initial velocity of 8 ft/sec. It crosses frictional patches as shown and slides down a ramp as shown. What is its final velocity?

In these problems keep your attention on the kinetic energy of the block. Just think: does the steop under consideration increase or decrease the KE? By how much? KE0 - Wf1 + PEg - Wf2 = KEf 1/2 mv02 - µ1mgl1 +mgh - µ2mgl2 = 1/2mvf2 (M'S CANCEL) 1/2 82 .3(32 * 2) + 32 * 5 - .4(32 * 3) = 1/2vf2 vf = 16.4 ft/sec

29. A mass m slides with velocity v_o over a frictional region of length L and coefficient of friction µ. How fast is it going on the other side?

KE0 - Wf = KE 1/2mv02 - µmgl = 1/2mv2 v2 = v02 - 2µgl v = "root"(v02 - 2µgl)

30. A 3 kg mass is pushed by a 7 nt force for 2 m, and then slides down a 3 m high ramp. How far will it skid before stopping on a 0.2 frictional surface?

W + PE = Wf 7 *2 + 3 * 9.8 * 3 = .2(3 * 9.8)x x = 17.4 m

31. Mass m is acted on by force F for a distance s. It then slides over frictional region 1, down the incline, and over frictional region 2. How far will it go before stopping?

W - Wf1 + PEg = Wf2 F * s - µ1mgL1cosØ + mgh = µ2mgx x = (Fs - µ1mgL1cosØ + mgh) / µ2mg

32. Mass m moves with velocity v_o, slides up the incline with frictional region as shown. How fast is it moving on the top?

KE0 - PE - Wf = KE 1/2mv02 - mgh - µmgcosØL = 1/2mv2 v = "root"(v02 - 2gh - 2 µglcosØ)

33. A mass m1 falls a distance h, dragging m₂ as shown. How fast will they be traveling if they started at rest?

PE0 - Wf = KE1 + KE2 m1gh - µm2gh = 1/2m1v2 + 1/2m2v2 = 1/2(m1 + m2)v2 v = "root"((2gh(m1 - µm2)/(m1 + m2))

So far we have considered only one form of potential energy gravitational potential energy. There are numerous other ways energy may be stored in nature: chemical potential energy, electrical potential energy, and mechanical potential energy to name a few.

At this time we will consider only one new form: potential energy of a stretched or compressed spring. To calculate the amount of energy stored is a bit trickier than it was for gravitational energy, because the force increases as the spring is stretched.

To properly understand the problem requires another look at graphing. If a 5 nt force is applied over 6 meters, 30 nt-m of work is done. Graphically this looks like:

Notice that the work done can be represented by the area under the curve.

A spring exerts a force F = kx which we may graph like this:

The work done, then, in stretching the spring a length x_o can be found by taking the average force times the distance:

So: work = (1/2 kx_o)(x_o) = 1/2 kx_o² which is also the area under the curve!!

Thus we have a new expression for the energy in a stretched (or compressed) spring:

PEs = 1/2 kx²

34. A spring of constant k - 280 nt/m is compressed 0.8 m. When released it pushes a 2 kg projectile. What is the final velocity of the projectile?

PEs = KE 1/2kx2 = 1/2mv2 1/2(280)(.8)2 = 1/2(2)v2 v = 89.61/2 v = 9.47 m/sec

35. If the spring shown is compressed 2.0 ft, how far will the block slide along the frictional surface before stopping?

PEs = Wf 1/2kx2 = µmgs 1/2(40)(2)2 = .3 (.5 * 32)s s = 16.7 ft

36. The weight shown rests on a frictionless surface in equilibrium position. It is stretched 5.3 cm and then released.
a) How fast is it traveling when it again passes the equilibrium position?
b) How fast was it traveling at 1.0 cm before the equilibrium position?

a) PEs = KE 1/2kx2 = 1/2mv2 1/2(55)(5.3)2 = 1/2(80)v2 v = 4.39 cm/sec b) PEs = KE = PE 1/2(55)(5.3)2 = 1/2(80)v2 + (55)(1.0)2 v = 4.32 cm/sec

37. A 3.4 kg sack of dog food traveling at 8.0 m/sec splats into a wall and stops. How much K.E. is lost? Where does the energy go?

1/2mv2 = 1/2(3.4)(8.0)2 1/2mv2 = 108 J This energy goes into heating the mass and wall. While Kinetic energy is lost, total energy is unchanged.

38. The two masses shown collide inelastically. How much kinetic energy is lost?

Momentum is always conserved: m1v1 + m2v2 = (m1 + m2)v 6*4 + 8 * (-2) = (6 * 8)v v = .57 1/2m1v12 + m2v22 = 1/2(m1 + m2)v2 + E E = 1/2 * 6 * 42 + 1/2 * 8 * 22 - 1/2(6 + 8)(.57)2 E = 61.7 J

39**. Prove that in a perfectly elastic collision, where no loss of kinetic energy occurs, the relative velocity before collision equals the relative velocity after collision.

(1) m1v1 + m2v2 = m1v11 + m2v21 m1(v1 - v11) = m2(v21 - v2) (2) 1/2m1v12 + 1/2m2v22 = 1/2m1v112 + 1/2m2v212 m1(v12 - v112) = m2(v112 - v22) m1(v1 - v11)(v1 + v11) = m2(v21 - v2)(v21 + v2) Divide (2) BY (1) [m1(v1 - v11)(v1 + v11)] / [m1(v1 - v11)] = [m2(v112 - v22)] / [m2(v21 - v2)] So v1 + v11 = v21 + v2 v1 - v2 = v21 + v11

40. The block is pressed 140 cm back against the spring and released. How fast is it moving after it has passed over the frictional region?

PEs - Wf = KE 1/2kx2 - µmgl = 1/2mv2 1/2(120)(1.4)2 - .2(4 * 9.8 * 5) = 1/2(4)v2 v = 6.26 m/sec

41. The spring is compressed a distance x, and the block released. How far from the bottom of the cliff does it land?

PEs PEg - Wf = KE 1/2kx2 + mgh1 - µmgl = 1/2mv2 So v = (kx2/m + 2gh1 - 2µgl)1/2 But s = 1/2at2 h2 = 1/2gt2 t = (2h2/g)1/2 Thus s = vt s = (kx2/m + 2gh1 - 2µgl)1/2 (2h2/g)1/2

Our discussion so far has centered on the amount of energy expended. What about the rate of energy use? A Honda engine can drive a Cadillac up a hill, but it will have to move very slowly. The difference is the power of the two engines.

Power equals energy per unit time: P = E/t

The units of power are, therefore, joule/sec (common name: watt), or ft-lb/sec, or erg/sec. (Another unit, the horse power is 550 ft-lb/sec, or 746 watt.)

42. How much power is required to lift a 23 nt weight 8 m up in 2.0 sec?

P = ÇE / Çt P = (23 * 8) / 2 P = 92.0 nt-m/sec P = 92.0 WATT

43. How much power is required to drag a box with a 17 lb force for 50 ft in 11 sec?

P = ÇE / Çt P = ((17lb)*(50ft)) / 11sec P = 77.3 ft-lb/sec

44. Herman Leftur can run 47 ft (vertical distance) upstairs in 6.8 seconds. If Herman weighs 193 lb, what power was he exerting to perform this feat? (express in ft-lb/sec and in horsepower)

P = ÇE / Çt P = (193 * 47) / 6.8 P = 1.33 * 103 ft-lb/sec * 1hp / 550 ft-lb/sec = 2.43hp

45. A 23 kg Schnauzer is to be shoved along a level surface whose coefficient of friction is 0.3. What power is required to move it along at 1.7 m/sec?

P = ÇE / Çt P = (F * s) / t P = F * (s / t) P = µmg * v P = .3 * 23 * 9.8 * 1.7 P = 115 WATT

46. A boat traveling at 3.0 m/sec encounters a constant drag force of 780 nt. What power is required to keep it moving?

P = ÇE / Çt P = F * s / t P = (780nt)*(3m) / 1sec P = 2.34 * 103 WATT

47**. A car encounters a drag force which increases with its velocity: F_d = kv. If the car has an engine that can put out power P, how fast can the car go at top speed? (express in terms of P and k)

P = F * s / t = F * v But F= kv So P = kv * v = kv2 v = (P/k)1/2