Work and Power

Objectives1. The student will investigate and understand

the interrelationships among mass, distance, force and time through mathematical and experimental process. Key concepts include work, power and energy (PH.5 g)

2. The student will understand that energy is conserved (PH.6a)

3. The student will investigate and understand that energy can be transferred and transformed to provide usable work. (PH.8a,b)

WorkWork is the force necessary to

move an object a distance.

WorkNeed to know!

Work = Force x distance moved

(If the force is in the same direction as the displacement)

Unit of WorkNeed to Know

Work is measured in newton-meters. A special name is given to this unit:

Joule (J) = 1 Newton-meter

Work is a scalar NOT a vector

“Working” outA man benches 585 lb (245 kg)The distance from his chest to theTop of the lift is .75 m. Find thework done by the teacher for one rep (up and down)Known: Distance (d) = .75 mweight lifted = mg = 245kg x 9.8m/s2

Fy = may

FLift - mg = 0

FLift = mg = (245 kg)(9.8 m/s2) = 2403 N

WLift = FLiftd = (2403 N)(.75 m) x 2 = 3605 J

mgd

.5FL.5FL

What is the unit of work?

1. Newton

2. Meter

3. Joule

4. Newton meter

5. 3 and 4

How much work is necessary to lift 10 kg 5m in the air?

1. 10 N

2. 50 J

3. 490 J

4. 4900 J

10 kg

5 m

PowerNeed to know

• Power is the rate at which work is done

P = average power

= work/time

= W/t

Unit: Watt(w) = Joule/sec (J/s)

Power Example: Running StairsA 70 kg student runs up a flight of stairs in 4.0s. The

vertical height of the stairs is 4.5 m. Estimate the student’s power output in watts

Know: mass = 70 kg; time = 4.0 s; y = 4.5 mThe work is done against gravity:W= Force x distance; where force = mgAnd distance equals vertical distance yWork = (mass) x (gravity) x (y)W = (70 kg)(9.8m/s2)(4.5m)W = 3087 Joules (J)P = Work/timeP = 3087 J/ 4.0 sP = 772 W (Recall 746 W = 1hp) = 1.03 hp

mg

y = 4.5

How much power is required to lift 10 kg, 5 m in the air in 10 s?

10 kg

5 m

1. 49 w

2. 490 J

3. 490 w

4. 4900 w

Power Example: Bench PressIf a teacher benches 245 kg (weight = mg = 2405N) 0.75 m

ten times in 25 seconds, estimate the power in his chest and arms

The teacher moves the weight (Force required = mg = 2405N) a total distance of 7.5 m (.75m x 10) so

Work (W) = (Force) x (distance) x (#repetitions)(Assume no work done bringing weight down)

Work = (2405 N)(.75 M)(10)W = 18,038 Joules (J)

Power = Work/timeP = (18,038J)/(25 s)

P = 721 W

Recall• If we applied a force to an object

• Work = Force x Distance

• Previously in our examples, the force aligned with the distance

Force

Distance

Force

BUT, what happens if the force and the distance are ….

NOT in the Same Direction

Distance

Force Force

If this is the case

• We must use the component of the force in the direction of the displacement

F F

displacement

ӨFcosӨ

Work = Force x Distance x cosӨNeed to Know

Bottom Line

• You can always use the equation

Work = (Force)(Distance)(cosӨ)

W=FdcosӨ

• Even if the Force is parallel to the displacement

Force

Displacement

Ө = 0o Cos0o = 1

Work Done by a Constant ForceNeed to Know

• Work: the product of the magnitude of the displacement times the component of the force parallel to the displacement

W = FdcosWhere F is the magnitude of the constant

force, d is the magnitude of the displacement of the object, and is the angle between the directions of the force and the displacement

Problem Solving Techniques1. FBD: Sketch the system and show the force that

is doing the work (as well as others that may be involved)

2. Choose an x-y coordinate system - direction of motion should be one

3. Determine the force that is doing the work4. Find the angle between the force doing the work

and the displacement 5. Find the work done: W=(Force)(distance)cos

d = 40m

Fp Fp

Example: Work done on a crate

Fp = 100 N = 37 degDetermine the work done by the force acting on the

crate

Wp = Fpdcos = (100 N)(40 m)(cos37) = 3200J

d = 40m

Fp Fp