Simple Concepts Involving Work and also Energy
Let be anunbalanced pressure applied to an object, and also let
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When q = 0, F is parallel tod, cosq = 1, W = Fd.
Units of W = Newton-meters (N-m) 1 N m = 1 Joule.
When q = 90�: F isperpendicular to d, cosq = 0, W= 0! Carrying a weight synchronizes to W = 0.
IF d = 0: Pushing versus an immovable object, W = 0.
Both cases body offers chemical power for muscles to exert these pressures (you get tired,need even more Twinkies to save going)--in regards to mechanical occupational performed: ZIP!
Work down lifting a box
Force not in very same direction as displacement:
Work maintaining a satellite in circular orlittle above earth:
NO job-related done by gravity to save it up. (What around Launch!!!!)
Work done in avoiding a car:
Mechanical Energy: conventional meaning = the ability to do occupational.
Moving objects have the right to perform job-related (bowling round dislocations pins; hammer pushes in nail; carcreams cow) energy of motion = kinetic energy.
Suppose a "bullet" of mass m relocating at vo mushes into a block of soft clay and experiences a constant pressure F(decelerating at a constant price, a).
The force required to slow down the bullet is F = ma, where a is the deceleration. Theoccupational done through the distance s, W = Fs = mas.
Throughout the deceleration, v2 = vo2 + 2as; or as = 1/2(v2 - vo2)
or Work done ON OBJECT =
The mechanical occupational done on the object = the readjust in kinetic energy;
If W is positive, KE increases; IF W is negative, KE decreases.
Suppose a auto of mass 1200 kg falls vertically a distance of 24 m (founding from rest;i.e., voy = 0).
(a) What is the work-related done by gravity on the car?
Fgrav = mg; Dy = 24 m; Force and displacement in very same direction (down).
Fgrav = Fnet because gravity is only force acting on vehicle.
(b) Find last velocity of vehicle.
Using consistent acceleration (g):
Using Work-Energy Theorem:
Plug in: v = 22 m/s.
Suppose F acting on a mass m relies on x; e.g., F = kx We divide xinto little bit increments, Dxi,where Fi is the average pressure over that interval:
Total work-related going from x1 tox2 = Wtotal = S FiDxi = location under curve!! (C word = integral of F from x1 to x2) and also DKE = Wcomplete.
For linear force, F= kx, Area under curve from 0 to xf =
A mass m is relocating in a right line at velocity vo. It comes right into contact through a spring with pressure consistent k. How much will certainly thespring compress in bringing the mass to rest?
A spring exerts F proportional to x in both compression and extension (for reasonablex).
Change in KE = KEf - KEi = 0 -
IF we usage object and compush spring this same distance (xf= xo) and let go, what is last KE and also v? Work done by F on mass, W= +
Change in KE = KEf - KEi = mvf^2 + 0. Because of this, mvf^2 = kxo^2. Sincexf = xo, we watch that: |
Object in movement -- has kinetic power. With or without motion, objects deserve to have PotentialEnergy -- "potential to execute work".
Examples: Object (Mass) at top of a building
Mass pumelted up against Compressed Spring
Water behind a dam
A attracted bow
For mechanical devices -- finest to think of this form of energy as "power ofposition".
For spring, we do work-related compressing the spring -- when compressed, spring prepared todeliver the power ago by accelerating a mass and also providing it KE.
When you compress the spring a distance x, the work done to compress the spring = ; thework the spring deserve to currently provide W = (Conservative Force) The spring is shelp to containpotential energy U = W = . When you release the spring, you deserve to deliver thisamount of energy to a mass m, so: mvf^2 = .
Gravity is a conservative force, also. Work done in lifting mass m a distance h = W= mgh. As such, for h > 0, U = W = mgh.
So Gravitational Potential Energy (in vicinity of Earths Surface) is Ugrav = mgh = mgy (up +)
Suppose you throw a ball right up via velocity vo. from yo = 0; set Uo = 0. Kcurrently that:
REMEMBER: holds for going up and going down!!!!
The change in the potential energy = readjust in KE. Gives |v| for any kind of allowed y. (Max yoccurs when v = 0: , as formerly calculated).
Anvarious other create of Eq. (1): Ufinal - Uinitial = KEinitial - KEfinal; we can collect"initial" and "final" so that:
Ulast + KElast = Uinitial + KEinitial
SUM = Total Mechanical Energy = E
HOLDS FOR ALL CONSERVATIVE FORCES. "initial" is whenever you begin theproblem, "final" is whenever you want an answer. NOTE that if ylast = yinitial,i.e., come earlier to wright here you started, E the exact same, independent of path.
Nonconservative forces - E counts on course. Best instance is frictional pressures. You dowork against them, yet you cant gain it ago (goes into heat).
Conservative: Uinitial + KEinitial = Ufinal+ KEfinal
Non-Conservative: Uinitial + KEinitial = Ufinal+ KEfinal + Q(heat)
Q represents shed mechanical energy; if we encompass, Energy still conoffered.
Suppose you throw 3 balls of equal mass m off a cliff via the exact same rate, vo, however at 3 angles relative to the horizontal presented infigure: 0�, �45�. Which sphere, 1, 2, or 3 strikes the ground (distance h below) with thebiggest speed? Neglect air resistance.
Answer: They all hit the ground via the very same speed. In all 3 cases, at throw,
Could be all various mass; same outcome.
Skier through Friction
A skier of mass 80 kg starts from remainder down a slope wbelow Dy = 110 m. The speed of the skier at the bottom of theslope is 20 m/s.
(a) Sexactly how that the device is nonconservative.
Given m, vo, vf, and h. Find Ei and also 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 x104 J
Because Ef i, device lost energy -- non-conservative.
(b) How much work is done by the non-conservative force?
Q = Ei - Ef = = 7 x 104 J (approximately 80% shed toheating up the snow).
Conservation of Energy holds for conversion between all kinds of energy: mechanical,electrical, thermal, nuclear, chemical. On microscopic / submicroscopic levels -- theseforms of energy are describable by kinetic and potential energy. E.g., energy of chemicalbond consists of the motion KE, attractive and also repulsive forces (therefore PE) fromelectrons and also nuclei.
In nuclear reactions -- need to take right into account that mass actually have the right to beconverted to energy: DE = Dmc^2, Einsteins famous equation.
Speed of a car raised by 50%. By what variable will certainly minimum braking distance beboosted (disregard reaction time)?
Braking force same. Therefore: Distance = 2.25 times original.
One car has 2x the mass of a 2nd automobile (=m), yet just fifty percent as much KE. When both carsincrease their rate by 5.0 m/s they have the very same KE. What were the original speeds ofthe cars?
First equation: v2 = 2v1
From second equation:
Power = price at which job-related is done (or price at which energy is transformed):
SI Units: Joule/s = 1 Watt
British: Horsepower; 1 hp = 550 ft lb/s = 746 watts.
If work-related is done by a consistent force, F:
Running the Stairs to the Stars (PhysicalSciences Bldg --Basement to 14th floor); Consider POWER.
1. Who could geneprice the greatest instantaneous power?
2. Who can geneprice the greatest average power?
What are "significant" output levels?
Human being in good physical shape -- 1/10 hp (75 W) at secure pace. O2 consumption -- 1 liter (1000 cm3)/minute.
Top athlete -- (lengthy distance sports-runners, skiers, bikers) 0.6 hp (~400 W); O2 consumption -- 5.5 liter/minute. Gossamer: (1979)humale powered aircraft, piloted by people class biker, crossed English Channel -- averaged190 W (0.3 hp).
FOR approx. 1 minute spurts - 450 - 500 watts.
For fraction of a 2nd -- numerous kW.
A 70 kg student runs up 2 flights of stairs (Dh = 7.0 m) in 10 s. Compute the students output in doing work againstgravity 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 expush elevator in Sears Tower (Chicago) averperiods a rate of 550 m/minute in itsclimb to the 103rd floor ( 408.4 m) above ground. Assume a full fill of 1.0 x 103 kg, what is the average power that the lifting motor mustprovide?
vavg = 550 m/minute x 1 min./60 s = 9.144m/s.
At consistent v, Force to lift = F = mg;
Pavg = Fvavg = (1.0 x 103 kg)(9.144 m/s) = 89.57kW = 90 kW.
(takes ~44 s to make trip).
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You want to loose weight; You therefore want to:
a)Run the stairs once/day as tough as you can, then hit the chips!, or
b) Sustain an task that burns ~ 1CALORIE/Minute practically eextremely day for 30-60 minutesand also dont hit the chips!
1 CALORIE = 1 Kilocalorie (4.186 kJ). 1 Kcal/minute = 70 Watts (substantialeffort). To loose weight need to exercise and diet!!!
Potential Energy of the Gravitational Field
On earths surchallenge, we take U = mgDy, wbelow Dy (or h) isthe height above the surconfront of the earth. This is an approximation bereason Fgrav depends on r (distance from center of earth). But Rearth >> Dy permits us to set Fgrav = to aconstant (mg).
For big distance alters, have to usage
Work done in going from R to r = Area under curve. WITH SOME EFFORT (INTEGRATE!!)can present this to be:
The minimum blast off rate to shoot a projectile off a world (or moon) and also never haveit fall earlier is called the escape velocity, vesc.
At the surface of the planet (r=R),
With no other pressures except gravity of earth, Emech stays continuous.
Consider projectile to arrive at r=8 via minimum KE (=0). Then Emech = 0 + 0!!! = Initial Emech.