Work and Power
Objectives
1. 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)
Work
Work is the force necessary to
move an object a distance.
Work
Need to know!
Work = Force x distance moved
(If the force is in the same direction as the
displacement)
Unit of Work
Need 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” out
A man benches 585 lb (245 kg)
The distance from his chest to the
Top of the lift is .75 m. Find the
work done by the teacher for one rep (up and down)
Known:
Distance (d) = .75 m
mg
2
weight lifted = mg = 245kg x 9.8m/s
d
Fy = may
FLift - mg = 0
.5FL .5FL
FLift = mg = (245 kg)(9.8 m/s2)
= 2403 N
WLift = FLiftd = (2403 N)(.75 m) x 2
= 3605 J
What is the unit of work?
1.
2.
3.
4.
5.
Newton
Meter
Joule
Newton meter
3 and 4
How much work is necessary to
lift 10 kg 5m in the air?
1.
2.
3.
4.
10 N
50 J
490 J
4900 J
10 kg
5m
Power
Need 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 Stairs
A 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 m
The work is done against gravity:
W= Force x distance; where force = mg
And distance equals vertical distance y
Work = (mass) x (gravity) x (y)
W = (70 kg)(9.8m/s2)(4.5m)
W = 3087 Joules (J)
P = Work/time
P = 3087 J/ 4.0 s
P = 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?
1.
2.
3.
4.
49 w
490 J
490 w
4900 w
10 kg
5m
Power Example: Bench Press
If 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/time
P = (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
Force
Distance
Force
If this is the case
• We must use the component of the
force in the direction of the
displacement
F
Work = Force x Distance
x cosӨ
F
Need to Know
Ө
FcosӨ
displacement
Bottom Line
• You can always use the equation
Work = (Force)(Distance)(cosӨ)
W=FdcosӨ
• Even if the Force is parallel to the
displacement
Force
Ө=
0o
Displacement
Cos0o = 1
Work Done by a Constant Force
Need to Know
• Work: the product of the magnitude of the
displacement times the component of the
force parallel to the displacement
W = Fdcos
Where 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 Techniques
1. 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 work
4. Find the angle between the force doing the work
and the displacement
5. Find the work done: W=(Force)(distance)cos

Fp
Fp
d = 40m
Example: Work done on a crate
Fp = 100 N
 = 37 deg
Determine the work done by the force acting on the
crate
Wp = Fpdcos = (100 N)(40 m)(cos37) = 3200J

Fp
Fp
d = 40m