Work, Power, & Energy

Energy is ability to do work.

• Work – a force applied to an object causes motion parallel to force direction. W = Fd cos

• No s = No W, = 90 = No W.• Neg work – F and s opposite (friction)

• W and E are Scalar quantities.

• Anytime E, is transformed E = W. E = W is an important equality frequently used for problem

solving.

ME and Internal E

• Mech E = KE, PE (grav), PE (elas)

• Int E (U) = heat, light, sound, chemical, etc.

• Generally some ME “lost” in interactions.

Units W = Fd = (N m)

kg m m . s2

kg m 2.s2

Joules (J)

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.

Before AfterEnergy = Energy

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

F

Ex 1: 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.

2. 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

KE and momentum

• KE = mv2

• 2

• Multiply top and bottom by m.

• m2v2.

• 2m

• p = mv.

• p2 = m2v2.

• Ek = p2/2m.

• Area under curve of F vs d • Work done or Energy

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.

IB Probs.

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

3. 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%

4. 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.

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

b. 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

Kerr pg 72 #2,3