Chapter 6: Work and Energy

Neither quantity is a vector!Woo hoo!

Work and Energy

• In physics, work = a force applied over a distance: W = Fd

• But if the force is not in the same direction, then you only care about that part of the force that is in the same direction

• So W = Fdcos(θ)• In physics, work and energy are equivalent• Both work and energy are measured in Joules• 1 joule = 1 Newton•meter

Work: Sample prob 1

• Kody Clegg is in the weight room doing a bench press. He lifts a total of 110 kg a distance of 0.8 m. His force is in the same direction the bar travels. What is the work Kody does on the bar?

• W = (110kg)(9.8m/s2)(0.8m) = 862 Joules• How much work is done as he lowers the bar?• W = (110kg)(9.8m/s2)(-0.8m) = -862 Joules• So what is meant by “negative work”?

Work: Sample prob 2

• Linda Monayong is pulling a piece of luggage. She pulls upwards at an angle of 500 from the horizontal with a force of 45 N and pulls the luggage a total of 75 m (the luggage moves horizontally along the floor). What is the work she has done on the luggage?

• W = Fdcos(θ) = (45N)(75m)(cos(500))• W = 2170 N

Work: Sample prob 3

• Some poor sap whose car (1000kg) won’t start is trying to push it, but it’s stuck. He pushes with a force of 200 N. What work is he doing on the car?

• 0 Joules (no distance moved)• David Green is standing on top of a railroad

flatcar, arms crossed and looking studly. David’s weight is 590 N. The boxcar travels 12 meters. How much work does David’s weight do on the boxcar as it travels?

• 0 Joules (force is 900 to direction of travel)

Kinetic Energy

• Kinetic Energy is the energy associated with motion.

• If an object has a mass M and is moving at a velocity V, then its kinetic energy is

• KE = ½ MV2

• Kinetic Energy is measured in Joules

Easy Kinetic Energy Sample Problems

• A 1100 kg car is traveling at 20 m/s. What is the car’s kinetic energy?

• ½ * 1100 kg * (20 m/s)2 = 220,000 Joules• A 50 gram bumblebee is flying at 0.75 m/s.

What is it’s kinetic energy?• ½ * 0.05 kg * (0.75)2 = 0.014 J• You get the idea

Work-Energy Theorem

• The work-energy theorem simply states that the amount of work you do on a system equals the change in kinetic energy.

• Work = ∆KE, where ∆KE = KEF – KEI, just like always.

• If the mass of the object doesn’t change (which is usually the case), then the change in kinetic energy is due to a change in velocity

• ∆KE = KEF – KEI = ½mVF2 – ½mVI

2

Work-Energy Problem

• A satellite of mass 5X104 kg is travelling at speed V0 = 1.1X104 m/s. Its engine exerts a force parallel with the direction of travel of a magnitude 4X105 N and pushes over a distance of 2.5X106 m. What is the final velocity of the satellite?

• Work = Fd, so W = (4X105 N)(2.5X106 m) = 1X1012 J

• Work = ∆KE, so KEF = W - KEI

Work-Energy Problem continued

• Work = ∆KE, so KEF = W - KEI

• ½ (5 X 104kg)(VF2) = 1 X 1012 J – ½(5X104kg)

(1.1X104m/s)2

• Solving for VF, we get VF = 1.27 X 104 m/s

Now, how did we solve that?

• First, we understand that the work done on a system is equal to the change in kinetic energy.

• Recall that Work = force * distance and solve for work done

• Then, recall that KE = ½ mv2 and recognize that ∆KE = KEF – KEI,

• We know enough info to solve for KEI

• All that is left is to solve for the unknown part of KEF

Work – Energy concepts

• A satellite is in a circular orbit around the earth and gravity is the only force acting on it.

• Does the gravity do any work? • Does the satellite’s KE change?

Work-Energy concepts continued

• Now consider the case of the earth’s orbit around the sun, which is actually an ellipse, not a circle.

• Does the sun do any work on the earth? Where? Where does it not do work?

• Why is this different From a circular orbit?

http://www.oglethorpe.edu/faculty/~m_rulison/Astronomy/Chap%2017/Images/ellipse.gif

Gravitational Potential Energy

• Force of gravity on an object (weight) is mg• Work done by gravity is that force * a distance• Let the distance over which weight acts be

height h• So work done by gravity W = mgh• Since work is equivalent to energy, then mgh

is equal to what we call the gravitational potential energy

Dr. Mason, we are intrigued. Tell us more about his “potential energy”, as you call it.

• Potential energy is stored energy that has not been used yet.

• Potential energy can be converted into other kinds of energy (notably kinetic energy).

• The important thing about potential energy is that it must be measured RELATIVE TO something.

• Gravitational PE depends on height and you have to measure the height from the ground.

• With no reference point, PE is meaningless

Falling stuff: Converting energy

• Let’s say a 3 kg rock is sitting at rest at a height of 10 meters above the ground. What is its speed just before it hits the ground?

• PE = mgh, so PE = (3kg)(9.8 m/s2)(10m) = 294 J• PE = ∆KE = KEF – KEI and KEI = 0 Joules (because

rock is initially at rest, so V = 0 m/s)• So PE = 294 J = KEF = ½ mVF

2

• 294 J = ½ (3kg)(VF2), so VF

2 = 196 J, so VF= 14 m/s

So how does that compare to the old way?

• How would we have found the speed of a falling object before?

• Easiest way was Vf2 = Vi

2 + 2AX and solve for Vf • And actually our new method reduces to that

equation readily (do you see it?).• Or we could have used F = ma, using a = Δv/Δt

and solving for time by using d = ½at2…• You get the idea. In general, energy

conservation is an easier approach

Conservative and Nonconservative Forces

• When talking about gravitational PE, we see that all that matters is the height above the ground.

• The object could fall straight down, or could follow a diagonal path, or whatever. It doesn’t matter. All that matters is the change in height.

• A force is conservative if the work done by it doesn’t depend on the path taken. Conversely, a nonconservative force is one whose work done does depend on the path taken.

Examples

• Conservative Forces– Gravity– Elastic springs– Electrical force

• Nonconservative forces– Friction– Air resistance– Tension– Normal Force

Concepts

• Let’s say you have a conservative force, like gravity. You fall from one height to the ground, then get raised back up to the original height. What is your net change in energy?

• Now consider a block sliding along a rough surface for 1 m. If the block slides back to its starting point, is the net change in energy = 0?

More Concepts

• Let’s say that you have a marble on a frictionless U-shaped track. It is released from a height H on one side. How high above the bottom does the marble roll on the other side?

• Now, add friction. Is the height reached greater or less?

More stuff

• Sometimes, both conservative and nonconservative forces act on a system (e.g. a block can slide under gravity down an rough incline).

• Remember that work and energy are equivalent, so we would have:

• mgh + WNC = ∆KE, or gravitational PE + work done by nonconservative force = change in KE

Conservation of Mechanical Energy

• We have introduced potential energy and kinetic energy.

• Total mechanical energy = PE + KE• E = mgh + ½ mv2

• If there are no nonconservative forces, then mechanical energy is conserved.

• EF = EI

• mghF + ½ mvF2 = mghI + ½ mvI

2

EF EI

More Conservation of Mechanical Energy

http://www.physicsclassroom.com/class/energy/u5l2b21.gif

So, Let’s say that in words

• The total energy stays constant.• As PE decreases (because height decreases),

KE increases• PE is being converted to KE• As PE goes down, KE goes up

Energy in a pendulum

http://serc.carleton.edu/images/sp/library/uncertainty/diagram_conservation_energy_si.jpg

Energy of a Pendulum• At position A, the

pendulum has only PE• Between A and B, it has a

mix of PE and KE (with PE decreasing and KE increasing)

• At B, it has purely KE• Between B and C, it has a

mix again (with KE decreasing and PE decreasing)

Sample Problem

• A motorcycle daredevil tries to leap across a canyon by driving horizontally off a cliff at 38 m/s. Ignoring air resistance, with what speed does the motorcycle hit the other side?

H0 = 70mH1 = 70m

Sample Problem continued

• mghF + ½ mvF2 = mghI + ½ mvI

2

• Every term has mass in it, so mass cancels out• ghF + ½ vF

2 = ghI + ½ vI2

• The only term that is unknown is VF, so solve for that.

• VF = vI2 + 2g(h0 – hF)

• VF = (38m/s)2 + 2(9.8)(70m-35m)• = 46.2m/s

√√

Now, see how much easier that was than before???

• Previously, we had to – Find the length of time the cycle was in the air

using the free fall equation– Use that info to find the final vertical speed– Use the Pythagorean theorem to combine

horizontal and vertical speeds to get total speed.

Another Sample Problem

• Tarzan is standing on a tree branch 8 m above the jungle floor. He is holding a vine that just happens to be 8m long and swings down. Assuming no air resistance and so forth, find his speed at the bottom of the swing.

• Can simplify equation because final height = zero and initial speed = zero

• Mgh = ½ mv2

• M(9.8m/s2)(8m) = ½ mv2

• V = 156.8 m2/s2 = 12.52 m/s√

Power

• As you have seen in the last lab experiment, power = energy/time

• Power also = work/time, since work and energy are equivalent

• W/time = Fd/t, but d/t = speed. So power also = Fv

• Units: 1 joule/sec = 1 watt• 1000 watts = 1 kilowatt• 1 horsepower (hp) = 745 watts

Power example

• An 1100 kg car, starting from rest, accelerates for 5 sec. the magnitude of acceleration is a = 4.6m/s2. what was the average power generated by the net force on the car?

• F = ma = (1100 kg)(4.6m/s2) = 5060 N• Vavg = ½(vf – v0) and vf = v0 + at = 23 m/s• Power = Fv, so P = (5060N)(11.5m/s) =

5.82X104 watts (78hp)