Simple Concepts Involving Work and Energy

Let wpe11.jpg (824 bytes)be an unbalanced force applied to an object, and let wpe12.jpg (796 bytes) be a resulting displacement. If F|| is the component of wpeF.jpg (839 bytes) along wpe13.jpg (796 bytes) , then the WORK done by wpeF.jpg (839 bytes), W, is given by W = F|| x d = dot product of the two vectors wpeF.jpg (839 bytes) and wpe13.jpg (796 bytes).

wpe14.jpg (9050 bytes)

When q = 0, F is parallel to d, cosq = 1, W = Fd.

Units of W = Newton-meters (N-m) 1 N m = 1 Joule.

wpe15.jpg (6497 bytes)

When q = 90: F is perpendicular to d, cosq = 0, W = 0! Carrying a weight corresponds to W = 0.

IF d = 0: Pushing against an immovable object, W = 0.

Both cases body uses chemical energy for muscles to exert these forces (you get tired, need more Twinkies to keep going)--in terms of mechanical work performed: ZIP!

Work down lifting a box

wpe16.jpg (24859 bytes)

Force not in same direction as displacement:

wpe17.jpg (10270 bytes)

Work keeping a satellite in circular orbit above earth:

        wpe18.jpg (9898 bytes)

NO work done by gravity to keep it up. (What about Launch!!!!)

Work done in stopping a car:

wpe19.jpg (15255 bytes)

Mechanical Energy: traditional definition = the ability to do work.

Moving objects can do work (bowling ball displaces pins; hammer pushes in nail; car creams cow) energy of motion = kinetic energy.

Suppose a "bullet" of mass m moving at vo mushes into a block of soft clay and experiences a constant force F (decelerating at a constant rate, a).

                        wpe1A.jpg (2791 bytes)

The force required to slow down the bullet is F = ma, where a is the deceleration. The work done through the distance s, W = Fs = mas.

During the deceleration, v2 = vo2 + 2as; or as = 1/2(v2 - vo2)

or Work done ON OBJECT = wpe1D.jpg (2251 bytes). (would get same result for non-constant F and a).

The mechanical work done on the object = the change in kinetic energy;   wpe1E.jpg (1634 bytes).

If W is positive, KE increases; IF W is negative, KE decreases. wpe1F.jpg (2282 bytes) -- often called Work-Energy Theorem (Net work done = change in kinetic energy

Suppose a car of mass 1200 kg falls vertically a distance of 24 m (starting from rest; i.e., voy = 0).

(a) What is the work done by gravity on the car?

Fgrav = mg; Dy = 24 m; Force and displacement in same direction (down).

wpe20.jpg (6245 bytes)

Fgrav = Fnet because gravity is only force acting on car.

(b) Find final velocity of car.

Using constant acceleration (g): wpe22.jpg (1504 bytes).

Using Work-Energy Theorem:

wpe23.jpg (6524 bytes)

Plug in: v = 22 m/s.

Variable Force

Suppose F acting on a mass m depends on x; e.g., F = kx We divide x into little increments, Dxi, where Fi is the average force over that interval:

wpe25.jpg (29033 bytes)  

Total work going from x1 to x2 = Wtotal = S FiDxi = area under curve!! (C word = integral of F from x1 to x2) and DKE = Wtotal.

For linear force, F= kx, Area under curve from 0 to xf = wpe26.jpg (1203 bytes).

                        wpe27.jpg (5207 bytes)

A mass m is moving in a straight line at velocity vo. It comes into contact with a spring with force constant k. How far will the spring compress in bringing the mass to rest?

A spring exerts F proportional to x in both compression and extension (for reasonable x).

wpe28.jpg (21813 bytes)

Change in KE = KEf - KEi = 0 - wpe2A.jpg (790 bytes) mvo^2. Work done by F on mass, W = -wpe2B.jpg (790 bytes) kxf^2. Therefore, kxf^2 = mvo^2 or wpe29.jpg (1687 bytes).

IF we use object and compress spring this same distance (xf = xo) and let go, what is final KE and v? Work done by F on mass, W = +wpe2D.jpg (790 bytes)kxo^2.

Change in KE = KEf - KEi = wpe2E.jpg (790 bytes) mvf^2 + 0. Therefore, mvf^2 = kxo^2. Since xf = xo, we see that: |wpe2F.jpg (813 bytes)| = |wpe30.jpg (824 bytes)|. Since we take a wpe31.jpg (804 bytes), does not give direction. Have to go to PICTURE; conclude that wpe32.jpg (1106 bytes).

Potential Energy

Object in motion -- has kinetic energy. With or without motion, objects can have Potential Energy -- "potential to do work".

Examples: Object (Mass) at top of a building

Mass pushed up against Compressed Spring

Water behind a dam

A drawn bow

For mechanical systems -- best to think of this type of energy as "energy of position".

For spring, we do work compressing the spring -- once compressed, spring ready to deliver the energy back by accelerating a mass and giving it KE.

When you compress the spring a distance x, the work done to compress the spring =   wpe33.jpg (1127 bytes); the work the spring can now deliver W = wpe33.jpg (1127 bytes) (Conservative Force) The spring is said to contain potential energy U = W = wpe33.jpg (1127 bytes).  When you release the spring, you can deliver this amount of energy to a mass m, so:  wpe2E.jpg (790 bytes) mvf^2  = wpe33.jpg (1127 bytes).

Gravity is a conservative force, also. Work done in lifting mass m a distance h = W = mgh. Therefore, for h > 0, U = W = mgh.

wpe34.jpg (11864 bytes)

So Gravitational Potential Energy (in vicinity of Earth’s Surface) is

Ugrav = mgh = mgy (up +)

Suppose you throw a ball straight up with velocity vo. from yo = 0; set Uo = 0. Know that: wpe35.jpg (1574 bytes).

REMEMBER: holds for going up and going down!!!!

MULTIPLY BY wpe36.jpg (935 bytes):

                        wpe37.jpg (4258 bytes)

Rearrange: wpe38.jpg (2783 bytes) (1)

The change in the potential energy = change in KE. Gives |v| for any allowed y. (Max y occurs when v = 0: wpeA.jpg (1686 bytes) , as previously calculated).

Another form of Eq. (1): Ufinal - Uinitial = KEinitial - KEfinal; we can collect "initial" and "final" so that:


Ufinal + KEfinal = Uinitial + KEinitial

SUM = Total Mechanical Energy = E

HOLDS FOR ALL CONSERVATIVE FORCES. "initial" is whenever you start the problem, "final" is whenever you want an answer. NOTE that if yfinal = yinitial, i.e., come back to where you started, E the same, independent of path.

Nonconservative forces - E depends on path. Best example is frictional forces. You do work against them, but you can’t get it back (goes into heat).

Conservative: Uinitial + KEinitial = Ufinal + KEfinal

Non-Conservative: Uinitial + KEinitial = Ufinal + KEfinal + Q(heat)

Q represents lost mechanical energy; if we include, Energy still conserved.

Suppose you throw three balls of equal mass m off a cliff with the same speed, vo, but at three angles relative to the horizontal shown in figure: 0, 45. Which ball, 1, 2, or 3 strikes the ground (distance h below) with the greatest speed? Neglect air resistance.

    wpe39.jpg (12915 bytes)

Answer: They all hit the ground with the same speed. In all three cases, at throw, wpe3A.jpg (1990 bytes). At collision with ground, wpe3B.jpg (1388 bytes).

Therefore,     wpe3C.jpg (4013 bytes)     independent of m!!

Could be all different mass; same result.

Skier with Friction

A skier of mass 80 kg starts from rest down a slope where Dy = 110 m. The speed of the skier at the bottom of the slope is 20 m/s.

         wpe3F.jpg (10358 bytes)

(a) Show that the system is nonconservative.

Given m, vo, vf, and h. Find Ei and Ef to test for conservation.

Ei = KEi + Ui = 0 + mgDy = (80 kg)(9.81 m/s2)(110 m) = 8.6 x 104 J

Ef = KEf + Uf = (1/2)mvf2 + 0 = (1/2)(80 kg)(20 m/s)2 = 1.6 x 104 J

Since Ef < Ei, system lost energy -- non-conservative.

(b) How much work is done by the non-conservative force?

Q = Ei - Ef = = 7 x 104 J (roughly 80% lost to heating up the snow).

Conservation of Energy holds for conversion between all types of energy: mechanical, electrical, thermal, nuclear, chemical. On microscopic / submicroscopic levels -- these forms of energy are describable by kinetic and potential energy. E.g., energy of chemical bond consists of the motion KE, attractive and repulsive forces (therefore PE) from electrons and nuclei.

In nuclear reactions -- must take into account that mass actually can be converted to energy: DE = Dmc^2, Einstein’s famous equation.

Driver’s Training

Speed of a car increased by 50%. By what factor will minimum braking distance be increased (ignore reaction time)?

                wpe40.jpg (6452 bytes)

Braking force same. Therefore: Distance = 2.25 times original.


One car has 2x the mass of a second car (=m), but only half as much KE. When both cars increase their speed by 5.0 m/s they have the same KE. What were the original speeds of the cars?

                    wpe43.jpg (6452 bytes)

First equation: v2 = 2v1

From second equation:

                    wpe42.jpg (6921 bytes)



Power = rate at which work is done (or rate at which energy is transformed):

        wpe46.jpg (5177 bytes)

SI Units: Joule/s = 1 Watt

British: Horsepower; 1 hp = 550 ft lb/s = 746 watts.

If work is done by a constant force, F:

wpe45.jpg (5541 bytes)

Running the Stairs to the Stars (Physical Sciences Bldg --Basement to 14th floor); Consider POWER.

1. Who could generate the highest instantaneous power?

2. Who could generate the highest average power?

What are "significant" output levels?

Person in good physical shape -- 1/10 hp (75 W) at steady pace. O2 consumption -- 1 liter (1000 cm3)/minute.

Top athlete -- (long distance sports-runners, skiers, bikers) 0.6 hp (~400 W); O2 consumption -- 5.5 liter/minute. Gossamer: (1979) human powered airplane, piloted by world class biker, crossed English Channel -- averaged 190 W (0.3 hp).

FOR approx. 1 minute spurts - 450 - 500 watts.

For fraction of a second -- several kW.

A 70 kg student runs up 2 flights of stairs (Dh = 7.0 m) in 10 s. Compute the student’s output in doing work against gravity in

(a) watts, (b) hp.

(a) W = FDh/t = mgDh/t = (70 kg) (9.81 m/s2) (7 m)/10s = 480 W

(b) W (in hp) = W (in W)/746 = 480/746 hp = 0.64 hp


The express elevator in Sears Tower (Chicago) averages a speed of 550 m/minute in its climb to the 103rd floor ( 408.4 m) above ground. Assume a total load of 1.0 x 103 kg, what is the average power that the lifting motor must provide?

vavg = 550 m/minute x 1 min./60 s = 9.144 m/s.

At constant v, Force to lift = F = mg;

Pavg = Fvavg = (1.0 x 103 kg)(9.144 m/s) = 89.57 kW = 90 kW.

(takes ~44 s to make trip).


You want to loose weight; You therefore want to:

a)Run the stairs once/day as hard as you can, then hit the chips!, or

b) Sustain an activity that burns ~ 1CALORIE/Minute almost every day for 30-60 minutes and don’t hit the chips!

1 CALORIE = 1 Kilocalorie (4.186 kJ). 1 Kcal/minute = 70 Watts (substantial effort). To loose weight have to exercise and diet!!!


Potential Energy of the Gravitational Field

On earth’s surface, we take U = mgDy, where Dy (or h) is the height above the surface of the earth. This is an approximation because Fgrav depends on r (distance from center of earth). But Rearth >> Dy allows us to set Fgrav = to a constant (mg).

For large distance changes, need to use  wpe48.jpg (2011 bytes).

wpe47.jpg (6816 bytes)

Work done in going from R to r’ = Area under curve. WITH SOME EFFORT (INTEGRATE!!) can show this to be:

wpe49.jpg (24031 bytes)The further m is from M, U becomes less negative, approaches largest value (0) at r=infinity.

wpe4A.jpg (13954 bytes)

wpe4B.jpg (11643 bytes)

Escape Velocity

The minimum blast off speed to shoot a projectile off a planet (or moon) and never have it fall back is called the escape velocity, vesc.

                                wpe4C.jpg (4749 bytes)

At the surface of the planet (r=R),

            wpe4D.jpg (3961 bytes)

With no other forces except gravity of planet, Emech remains constant.

Consider projectile to arrive at r=8 with minimum KE (=0). Then Emech = 0 + 0!!! = Initial Emech.


wpe4E.jpg (12467 bytes)

      wpe4F.jpg (11501 bytes)