WORK & ENERGY So far, we’ve been obsessed with motion and how it relates to Newton’s Laws and forces. Now we will be concerned with two scalar quantities, work & energy and how they are always conserved or remain constant. 6.1 WORK DONE BY A CONSTANT FORCE “Work” had a variety of meanings in every day language. It means something VERY specific in a physics class! ! work-- “a force acting through a distance” OR the product of the magnitudes of the displacement times the component of force parallel to the displacement. 2 W = F d cos è 2 ! F is the component of constant force F parallel to the displacement d. ! è is the angle between the direction of the F & d when F is in the direction of work, è = 0 AND cos 0 = 1 so . . . W = F × d ! Work is scalar and thus has only magnitude. ! UNIT--the unit of work is the N C m OR joule (J) ! Since W = Fd an object at rest has zero displacement, therefore W = 0 ! When a F is perpendicular to motion like the figure at the left, W = 0 even though his arms may get mighty tired and he may even become winded or break a sweat! ! W = Fd cos è ! Since the displacement is 90E to the motion; cos 90E = 0 and no work is done! ! Walk with constant velocity and NO work is done since a = 0, F = 0. ! When you begin or slow to stop there is a change in velocity, a 0 and work is done. ! This means you must specify by and on when discussing the forces as well as singling out the forces you are referring to. Often we use the NET force. Rene’ McCormick, NMSI Work & Energy 1
Transcript
WORK & ENERGY
So far, we’ve been obsessed with motion and how it relates to Newton’s Laws and forces. Now we willbe concerned with two scalar quantities, work & energy and how they are always conserved or remainconstant.
6.1 WORK DONE BY A CONSTANT FORCE
“Work” had a variety of meanings in every day language. It means something VERY specific in aphysics class!
! work-- “a force acting through a distance” OR the product of the magnitudes of the displacementtimes the component of force parallel to the displacement.
2 W = F d cos è
2 ! F is the component of constant force F parallel to the displacement d.
! è is the angle between thedirection of the F & d when F is inthe direction of work, è = 0
ANDcos 0 = 1 so . . . W = F × d
! Work is scalar and thus hasonly magnitude.
! UNIT--the unit of work is the N C m OR joule (J)! Since W = Fd an object at rest has zero displacement, therefore W = 0
! When a F is perpendicular to motion like the figure at the left, W = 0 eventhough his arms may get mighty tired and he may even become winded orbreak a sweat!
! W = Fd cos è! Since the displacement is 90E to the motion; cos 90E = 0 and no work is
done!! Walk with constant velocity and NO work is done since a = 0, F = 0.! When you begin or slow to stop there is a change in velocity, a � 0 and
work is done.! This means you must specify by and on when discussing the forces as well
as singling out the forces you are referring to. Often we use the NET force.
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Example 6-1
pA 50 kg crate is pulled 40 m along a horizontal floor by a constant force exerted by a person, F = 100 N,
fwhich acts at a 37E angle as shown. The floor is rough and exerts a friction force F = 50 N. Determinethe work done by each force acting on the crate, and the net work done on the crate.
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! Friction--can do NEGATIVE work. It is negative when the F or its F2 component acts in thedirection OPPOSITE to the direction of motion.
Example 6.2
a) Determine the work a hiker must do an a 15.0 kg backpack to carry it up a hill ofheight h = 10.0 m.
b) Determine the work done by gravity on the backpack
c) Determine the net work done on the backpack. For simplicity, assume the motion is smooth and at aconstant velocity.
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Example 6-3
The Moon revolves around the Earth in a circular orbit, kept there by the gravitational force exerted bythe Earth. Does graveity do positive, negative or no work on the moon?
WORK DONE BY A VARYING FORCEWhat if the Force applied is not constant? What if it keeps changing?
Consider a rocket launch. The work done varies since gravitational attraction varies as theinverse square of the distance from the Earth’s center.
2! WORK can be determined by the area under the curve of a plot of F (Fcos è indirection of motion) vs. d [don’t use just any old plot!]
2! Ä W = F Äd! OF COURSE, the work done between any two points on the curve is the area under
the curve BETWEEN those two points.
KINETIC ENERGY AND THE WORK-ENERGY PRINCIPLE
! Energy--the capacity to do work. There are many types of energy. Most fit into two broadcategories: Potential Energy or Kinetic Energy
! Law of Conservation of Energy-- Energy cannot be created nor destroyed, just change form.a.k.a. The First Law of Thermodynamics" Changes are NOT perfect--once we realized the heat lost was it’s own form of energy the
conservation of TOTAL energy made much more sense!" Friction also usually accompanies these changes--still lost as heat (thermal energy)
! Kinetic Energy--a moving object can do work on another moving object it strikes and can thusmove that object, therefore W is done. This means it fits the definition for energy and is usuallythought of as the “energy of moving objects”. [Remember the average KE of molecules istemperature.]
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! DERIVATION OF ½ mv2
1Consider an object of mass m moving in a straight line with an initial speed v . To accelerate it
2 netuniformly to a speed v , a constant net force F is exerted parallel to its motion over a distance d.
net netW = F d
Apply Newton’s 2nd Law
netF = ma
From Kinematics
2 1v = v + 2ad and solve for a2 2
2 1 neta = v - v substitute into F = ma to determine the work done2 2
2d
the d’s cancel and you have
net 2 1W = ½ mv - ½ mv = Ä KE2 2
WORK- ENERGY PRINCIPLE OR THEOREM:
netWe define translational KE to be ½ mv AND W = Ä KE2
VALID ONLY FOR NET FORCE AND NET WORK!!! NOTE: KE is % to mass AND KE is % to v2
" double the mass, KE doubles" double the velocity, KE quadruples
Example 6-4
A 145 g baseball is thrown with a speed of 25 m/s.
a) What is its KE?
b) How much work was done on the ball to make it reach this speed, if it started from rest?
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Example 6-5
How much work is required to accelerate a 1000 kg car from 20 m/s to 30 m/s?
Example 6-6
An automobile traveling 60 km/h can brake to a stop within a distance of 20m. If the car is going twoceas fast, what is its stopping distance? The maximum braking force is approximately independent ofspeed.
6.4 POTENTIAL ENERGY! Potential Energy--the energy of position; many kinds including gravitational and that in a
wound or compressed [or stretched] spring.! DERIVATION OF POTENTIAL ENERGY
In order to lift an object of mass m vertically an upward F @ least equal to mg
1must be exerted on it say, by a hand. Lift without acceleration a height h, from y
2to y , a person does W equal to the product of the needed external force
extF = mg upward a vertical distance h
ext extW = F d cos è = mgh
2 1 =mg (y - y )
Gravity is also acting and does work equal to
g g g W = F d cos è = mgh cos 180E (F and d point in opposite directions)
g W = -mgh
2 1 = - mg (y - y )Allow the object to free fall under the action of gravity and it acquires a velocity
v = 2gh AND after falling h2
KE = ½ mv = ½ (2gh) = mgh2
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Should it strike another object it can do work W = mgh (W-E Principle)
Gravitational Potential Energy PE = mgh
The higher an object is, the more gravitational PE it has.
ext 2 1W = mg (y - y )
ext 2 1 gW = PE - PE = ÄPE and W = -ÄPE
! It can be confusing as to where to set “ground zero”. It is a matter of convenience since it is theÄ in vertical h that matters [beware inclines!]
Example 6-7
A 1000 kg roller-coaster car moves from point A to point B and thento point C.
a) What is its gravitational PE at B and C relative to point A?
b) What is the change in PE when it goes from B to C?
c) Repeat parts a) and b), but take the reference point to be at point C.
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! A ÄPE associated with a particular force is equal to the negative of the work done by that force ifthe object is moved from one point to a second point.
! Elastic PE--a spring (or other elastic object) has PE when compressed (or stretched) for whenreleased it can do work [that’s why they call it potential--it has the potential to do work!]
! For a person to hold a spring either stretched or compressed an amount x
pfrom its normal length requires F % x so, can you feel it coming?
pHooke’s Law F = kx
! k is the spring constant and is a measure of stiffness" the spring exerts a F in the opposite direction according to
Newton’s 3rd law
s" F = -kx & is sometimes called the restoring force! To calculate the PE of a stretched spring:
p W = F x = ½kx(x) = ½ kx2
Elastic PE = ½ kx2
PE is an attribute of a system and has no neat, clearformula.
It depends upon the system.
6.5 CONSERVATIVE AND NONCONSERVATIVE FORCES
Work done against gravity does NOT depend on a path taken. Itsimply depends on Äh as in mgÄh.
! Conservative forces--independent of the path taken suchas PE = mgh. Only the Ä h matters.
! NONconservative forces--friction for example--doesdepend on the path taken; W = Fd cos è and if d increasesso does the work!
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! Potential Energy can be defined only for a conservative force.! Extend the W-E Principle to include PE
netW = G W done by conservative and nonconservative forces
net c ncW = W + W
netAND W = Ä KE so.....
c nc 2 1W + W = Ä KE AND Ä KE = KE - KE
nc cW = ÄKE - W
c cW can be written in terms of PE as W = -ÄPE
SUBSTITUTE and you get
W-E Principle in its general form:
ncW = Ä KE + Ä PE
The work done by NC forces acting on an object is equal to the total change in potential and kineticenergies. ALL THE FORCES ACTING ON A BODY MUST BE INCLUDED in either the PE (if a
ncconservative force) or in W term on left but NOT in both!
6.6 MECHANICAL ENERGY AND ITS CONSERVATION
conservedIf only F are acting on a system life is simple:
ncW = 0 AND
Ä KE + Ä PE = 0 OR
2 1 2 1KE - KE = PE - PE
! Now we define a quantity E, called the total mechanical energy of our system as the sum of theKE and PE at any moment
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E = KE + PE
2 2 1 1can re-write KE + PE = KE + PE
2 1or E = E = constant which means it is a conserved quantity.
! Principle of Conservation of Mechanical Energy--If only conservative forces are acting, thetotal mechanical energy of a system never increases nor decreases in any process. It staysconstant--it is conserved.
6.7 PROBLEM SOLVING USING CONSERVATION OFMECHANICAL ENERGY
Falling rock. At moment of release only PE = mgh is important. During fall,PE decreases with height and KE increases with the square of the velocity so Eremains constant @ any point along the path.
E = PE + KE = mgh + ½ mv2
1 1 2 2AND ½ mv + mgh = ½ mv + mgh2 2
Just before the rock hits the ground ALL the PE º KE
0 + mgh = ½ mv + 02
Example 6.8
If the original height of the stone in fig 6-17 is h = 3.0 m, calculate ht stone’s speed when it has fallen to1.0 m above the ground.
ENERGY BUCKET
A simple way to visualize energy conservation is with an “energybucket”. The total E remains the same throughout the exercise.
Just before impact, the speed = (2gh)½
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Example 6.9
Assuming the height of the hill in fig 6-19 is 40 m, and the roller-
coaster car starts from rest at the top, calculate
a) The speed of the roller-coaster car at the bottom of the hill
b) at what height it will have half this speed.
Example 6.10
Two water slides at a pool are shaped differently but start at thesame height. Two riders, Paul and Kathleen, start from rest at thesame time on different slides.
a) Which rider is traveling faster at the bottom?
b) Which rider makes it to the bottom first? Ignore friction.
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gPole Vaulting--KE runner º elastic PE of pole º PE º pole straightens (restoring force) º all E is
gnow in PE . Pole is device for storing energy!
Example 6.11
Estimate the KE and the speed required for a 70 kg pole vaulter to just pass over a bar 5.0 m high. Assume the vaulter’s center of mass is initially 0.90 m off the ground and reaches its maximum height atthe level of the bar itself.
For springs or other ELASTIC substances [only!]
1 1 2 2 ½ mv + ½ kx = ½ mv + ½ kx 2 2 2 2
Example 6.12
A dart of mass 0.100 kg is pressed against the spring of a toy dart gun. The spring (k = 250 N/m) iscompressed 6.0 cm and released. If the dart detaches from the spring when the latter reaches its normallength (x=0), what speed does the dart acquire?
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Example 6.13
A ball of mass m = 2.60 kg, starting fromrest, falls a vertical distance h = 55.0 cmbefore striking a vertical coiled spring,which it compresses an amount Y = 15.0cm. Determine the spring constant of thespring. Assume the spring has negligiblemass. Measure all distances from thepoint where the ball first touches theuncompressed spring. (y = 0 at this point)
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Example 6-14
Dave jumps off a bridge with a bungee cord tied around his ankle. He falls for 15 meters before thebungee cord begins to stretch. Dave’s mass is 75 kg and we assume the cord obeys Hooke’s law,
F = -kx, with k = 50 N/m. If we neglect air resistance, estimate how farbelow the bridge Dave will fall before coming to a stop. Ignore themass of the cord.
6.8 OTHER FORMS OF ENERGYBesides the general and broad categories of PE & KE there’s! electric! nuclear! radiant! thermal! chemical! mechanical! and more.....
! Transfer of E is accompanied by performing work. Work is done when E is transferred from oneobject to another
6.9 ENERGY CONSERVATION WITH DISSIPATIVE FORCES
We’ve been neglecting air resistance, friction occassionally, etc.! truth: dissipative forces are those that keep mechanical E transfers from being perfect
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! Consider heat as thermal energy and the law of cons. of E holds true.
! Apply the W-E Principle to things like the frictional force:
ncW = Ä KE + Ä PE
nc fW = -F d (F and d are in opposite directions hence the negative sign)
f 2 1 2 1-F d = ½ mv - ½ mv + mgh - mgh OR2 2
1 1 2 2 f½ mv + mgh = ½ mv + mgh + F d2 2
Example 6.15
The roller-coaster car in example 6-9 is found to reach a vertical height of only 25 m on the second hillbefore coming to a stop. It traveled a total distance of 400m. Estimate the average friction force on thecar, whose mass is 1000 kg.
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6.10 POWER
! average power--rate @ which work is done
P = Work measured in watts 1 watt = 1 J/s = NCm/s
time
! horsepower--equal to 746 watts
Example 6.16
A 70 kg jogger runs up a long flight of stairs in 4.0 s. The vertical height of the stairs is 4.5 m
a) Estimate the jogger’s power output in watts and horsepower.
b) How much energy did this require?
Example 6.17
The Nova laser at Lawrence Livermore National Lab has ten beams, each of which has a power outputgreater than that of all the power plants in the US. Where does this power come from?
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Example 6-18
Calculate the power required of a 1400 kg car under the following circumstances:
a) The car climbs a 10E hill at a steady 80 km/h
b)The car accelerates along a level road from 90 to 110 km/h in 6.0 s to pass another car. Assume aretarding force on the car of 700 N throughout. (This force is more about air resistance than friction)
