Work and Energy

Work, Power, & Energy

Energy offers an alternative analysis of motion and its causes.

Energy is transformed from 1 type to another in interactions between objects when work gets done.

Anytime energy, E, is transformed E = W.

Def: Energy is ability to do work.

Work is done when force applied to an object or particle causes motion parallel to force direction.

Work and Energy are Scalar quantities.

If force is at an angle to d, then W = F x d (cos )

F = applied force parallel l l to displacement Newtons.

d = displacement caused by force – meters

= angle between force and displacement.

Work can be zero even if force is applied.

W = Fd cos

• The work done by the force F is zero if:

– d = 0 m (no displacement)

– cos = 0 when = 90° (force perpendicular to the displacement).

• Zero Work

• The gravitational force between the earth and the sun does not do any work !

• Negative Work:

• Force in opposite d to displacement.

Units W = Fd = (N m)

kg m m . s2

kg m 2.s2

Joules (J)

Ex 1. A student lifts a 15-N -kg box 2-m straight up.

• a. How much work did he do on the box? • F Same direction as d ( = 0°.)• W = Fd cos = (15-N)(2-m) cos 0° = 30 J.

• b. How much work did gravity do on the box?• W = Fd cos = (15-N)(2-m)cos 180° = - 30J.

Whenever work is done on or by an object, its energy changes. W = E.

• Work changes an object’s energy from one form to another, increases or decreases it.

• Common Mechanical E

• Ep = mgh.

• Ek = ½ mv2.

F

Ex 2: A weight w, of 4-kg sits on a 30o ramp of length L = 10-m. If the work done to slide the weight to the top of the ramp was 300-J, how much work was done against friction?

Work done moving W, 10-m up ramp transforms E (gain in PEg). Ignoring friction, work to pull weight:

F

W = Fdll = F x L (length)

But pulling F = mg sin .

Or , W = PEg = mgh.

Sin 30o = h/L , h = sin 30x10 m = 5 m

PE = 4x10 x5 = 200-J w/o fric

• 300-J – 200-J = 100-J to overcome Ff.

F = 4x10 sin 30 = 200-J to pull weight w/o fric.

Types of Energy

Kinetic (KE) – energy of motion

Gravitational Potential (GPE) – energy due to height in gravitational field.

Elastic Potential – energy stored in shape deformations.

Mechanical Energy ME

Chemical- energy stored in chemical bonds.

Electrical Energy – due to charge separation.

Internal or Heat Energy (Q) – due to vibration of atoms.

Light / Radiation – atomic vibrations.

Mass/Nuclear Energy = Stored between subatomic particles. E = mc2.

More…

Internal Energy U, Q

Energy is conserved quantity!

The total of all energy types at one point in time = the total of all the energy types at any other point in time in a closed system.

E1 = E2

KE and momentum

• KE = mv2

• 2

• Multiply top and bottom by m.

• m2v2.

• 2m

• p = mv.

• p2 = m2v2.

• Ek = p2/2m.

IB Probs.

Potential Energy – Stored Energy

An object can store energy as the result of its position.

Elastic PE

Elastic materials store energy when stretched or compressed.

• Work done by non-constant force.• Area under curve of F vs d = work done• W = Fd = ½ kx2.

Work done non-constant force. E/W = area under curve for F vs. x graph for spring. PE s

Area = WW = ½ F x

But F = kx so

W = ½ kx x

E = ½ k x2.

Elastic PE

• PE elastic = ½ kx2. – x is the dif btw relaxed length and new

length in meters sometime x.

– k is a constant for the material N/m.

– PE in joules.

Mechanical Energy & Non-mechanical Energy

Mechanical Energy =KE, GPE, EPE.

Non-mechanical/ Internal = heat, light, chemical, nuclear, magnetic, electric, sound.

Conservation of Energy during transformationBefore After

Energy = Energy

W + KE + PEg + PEs before = KE + PEg + PEs + U.

Conservation of Energy

Energy Transformations What are the energy transformations

in the following situations?

• Bouncing a ball on the floor.

• Dropping a weight suspended from a spring.

• Driving a car into a brick wall.

Calculations Using Conservation of Energy

• Identify the energy transformations.

• Set up equalities.

• Break down equations if necessary.

Ex. An archer uses a bow with a constant of 96 N/m. He stretches the bow back 50 cm and releases a 0.2 kg arrow. What is the velocity of the arrow upon leaving the bow?

E = E

•PEs = KE

•½ kx2 = ½ mv2.

•½(96)(0.5)2 = ½ (0.2)v2.

•11m/s

Ex 3. A skier starts from rest at the top of a 45 m high hill. He skis down the hill which is at a 30o incline, down into a valley, and coasts 40m up to the top of the next hill. Assume no friction.

• How fast is he going at the valley bottom?

• What is his speed at the top of the 2nd hill?

How fast is he going at the valley bottom?

What is his speed at the top of the 2nd hill?

Pendulum Problems

PE to KE to PE

Compare the PE & KE at the top & bottom of the swing.

• Top KE =0

• PEg maximum

• Bottom of swing PEg = 0

• KE is maximum

Energy Lost to Friction (or other internal forms)

Work done = E of an object. If some E is “missing” after your calculations are done, it was probably “lost” to internal energy.

The “lost” energy can be the difference between the sum of the initial work, energies and the final energies.

Set up a work-energy transformation equation to solve.

Ex 5. A girl pushes a box at a constant speed 10-m by doing 15-J of work. How much work was done by friction?

• W = E.• W = KE + heat gained. • KE = 0• 15 J

Ex 6. A 5 kg box slides down a hill from a height of 6-m. If its KE at the bottom of the hill is 200-J, how much energy was lost to friction?

• PE = mgh should equal gain in KE

• =(5kg)(9.81m/s2)(6m)

• =294 J but KE only 200J

294 J – 200 J = 94 J Lost

Energy & Work Something for Nothing?

Work done against gravity is conservative!

If a box gains 4 J of PE, it takes 4 J of work to give it the PE regardless of whether you lift it straight up or push it up a (frictionless) ramp!

Consider sliding a box up a ramp to change its PEg.

The force you apply is to overcome the weight.

(no friction).

As the ramp angle gets lower the distance to raise the object must get longer to attain the same PE.

For the same change in PE, the distance can be shorter or longer. But work Fd is the same.Either large F & short d, or small F & long d.

h2h1d1

d2

larg

e F

small F

h1 h2

PowerDef: Rate at which work is done.

Power = work/time = W J Nm

t s s

1 J/s = 1 Watt (W)

If it takes 10 seconds to do 50 J of work, then

P = 50J10s

5 J/s or 5W.

Power is related to speed that work gets done.Watts are commonly used to describe how fast motors & engines will do work.

P = W Fd Fvt t

1 watt is the rate at which a 1N forcedoes work if it moves a body 1m/s.

1 horsepower (HP) = 746 watts.

Efficiency

Usually, some energy is lost to heat, light, sound, etc. We must do extra work.

Efficiency is a percentage:

useful work done x 100%

actual work done

Ex 9: An engine with a power output of 2.0 kW pulls an object with weight of 1000 N at a constant speed straight up. A constant friction force of 300-N acts on the engine. The object is lifted a distance of 8.0 m.

a. Find the speed of the object .

P = Fv. The engine must do work against weight (1000-N) and friction ( 300N), so Fnet = 1300N.

2000 W = ( 1300 N)v.

1.5 m/s = v.

b. Find the efficiency of the engine.

useful work done x 100%• actual work done

•Useful work = mgh = (1000-N) (8m)•= 8000 J•Actual work = Fd = (1300-N) (8m)•10,400 J•8000/10,400 x 100% = 77%

Ex 2: An engine with a power output of 1.2 kW drags an object with weight of 1000 N at a constant speed up a 30o incline. A constant frictional force of 300 N acts between the object and the plane. The object is dragged a distance of 8.0 m.

1. Find the speed of the object and the efficiency of the engine.

2. calculate the energy output per second of the fuel used.

Speed of the Object

• The downhill weight component is:• 1000 N (sin 30o) = 500 N.• The total force the engine must overcome is

weight + friction:• 500 N + 300 N = 800 N.• P = Fv v = P 1200 W

F 800 N• v = 1.5 m/s.

• The machine lifts the weight to a height: 8.0m (sin30o) = 4.0 m.

• The useful work is (mg)h:• (1000 N)(4.0 m) = 4000 J.• The actual work done is: (800 N)(8.0m)= 6400 J.• Eff: useful work x 100%

actual work

• 4000 J x 100% • 6400 J

• =63 %

You can calculate the energy output per second of the fuel used.

Since the machine has a power output of 1.2 kW and is 63% efficient, the fuel must produce energy at a rate of:

1.2 x 103 J/s = 1.9 x 103 J/s. 0.63

IB Question

The universe season 2 Gravity, Roller Coasters, Space Travel & PEg.

http://www.youtube.com/watch?v=Hsv9E8tMGe4