1.
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Work Energy & Power
Keith Warne
x
Fx
W = F • x cos
2.
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m mm
Energy – Kinetic Energy
Energy is the capacity to do
work.
Energy is a scalar quantity
and has the same unit as
work viz. Joule.
When ENERGY is
transferred WORK is done.
Energy of movement
W = f.x
F = m.a
=>W = max
vf
2 = vi
2 +2ax but
vi = 0 and vf = v
=> x = v 2 /2a
W = m.a.(v 2/2a)
=> EK = 1/2mv 2m a m.s-2FORCE
A mass (m) is accelerated from rest
to a velocity v.
Vi = 0 vf = v
m
3.
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Work Energy Theorem
The work done on an object by a net force is equal to the
change in the object’s kinetic energy:
Ek = work done (Net)
If V = const then Ek = Wnett = ........
With Conservative forces (...............................) at any
point during the fall of an object the mechanical energy
is equal to the potential energy of the object before it
began to fall.
E (mech) = .........
Ep (........) = Ek (.............)
With Non conservative forces present (friction):
Energy is lost through work then
E = Ep + Ek + ...........................
Wnett = Wf1 + Wf2 + etc.
EK = 0
Ep = mgh
Ek = (mgh)
Ep = 0
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4.
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An object is lifted vertically by a force F at a constant velocity.
2 kg
3 m
F WF = F.x cos
= (- Fg).x cos = -(2*-9.8)*(3)*(1)= 58.8 J
Wg = F.x cos
= Fg.x cos (180) = (2*9.8)*(3)*(-1)= -58.8 J
Work Energy Theorem
Ek = work done (Net) = 0 (const v)
Work done by F + Work done by g = 0
Wf = -Wg
.: Fnett = 0 .: F + Fg = 0 .: F = -Fg
Fg
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An object is pushed up a slope by a force F at constant velocity.
Frictionless slope.
3 m
Fnett = 0 (const v) .: |F//| = |F|
WF = F.x cos
= F// x cos
= (2*9.8*sin30)*(3)*cos(0)
= 29.4 J
Wg = F.x cos = (2*9.8)sin30 *(3)cos (180) = - 29.4 J
Wg = F.x cos = Fg .H cos (180) = (2*9.8)*3*sin(30)*(-1) = - 29.4 J
Work Energy Theorem (Wnett )
Wnett = Ek = 0 (const v.)
30o
Fg
F// = Fg sin 30
= 2*9.8 sin 30
H = 3sin30
6.
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A pendulum
• A person of 60kg is lifted to a height of 30m on a slingshot
pendulum and then released what is its maximum speed?
m
h
ETop = Ep + EK
= mgh + 1/2mv2
= (60)(10)(30) + 0
= 18000J
Total Mechanical Energy
(Top)
EBottom = Ep + Ek
= mgh + 1/2mv2
18 000 = 0 + 1/2 (60)v2
v2 = 18000/30 = 600
v = 600 = 24.5 m.s-1
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7.
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Power
• Power is the RATE at which WORK is done.
Power =
Units: Watts (W) =
Work
time
Joules (J)
Time (s)
Since v = x/t, Power can be found by
P = F.v
If a force of 20N is exerted over a distance of 5m for a time of 30s the power
used would be. W = F.x = (20).5 = 100J P = W/t = 100/30 = 3.3 W
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8.
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WE Th Eg2
• The horizontal force F of 4 N is pulling the block whilst a frictional force
f of 2 N opposes the motion. If the block’s velocity is 2 m.s-1 when it
passes point S, what is its velocity when it passes point D? How much
work is done against friction?
Fnet = F + f .: Fnett = 4 – 2 = 2 N
and the angle between Fnet and ∆x is 0o
Wnet = ∆Ek Fnet∆xcos = ∆Ek
F∆xcos0o = ½ mvf
2 – ½ mvi
2
2x10x1 = ½ 2vf
2 – ½ 2x22 = vf
2 – 4
vf
2 = 20 + 4 = 24
vf = 4.90 m.s-1
m=2kg
F = 4 N
S 10 m D
f = 2 N
Wf = Ff∆xcos
= 2*10*cos(180)
= 20*(-1)
= -20 J
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9.
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Note on Worka) Work done by 680N force:
WF = F.Δxcosθ W = 680 x 80 x cos0 = 54400 J
b) The work done on the object by the gravitational force.
Fg = mg = 40 x 9,8 = 392 N
Wg = F.Δxcosθ = 392 x 80 x cos180 = -31360 J
c) The net work done on the object.
Wnet = WF + Wg = 54400 – 31360 = 23040 J
d) The gain in potential energy
U = mgh = 40 x 9.8 x 80 = 31360 J
e) The gain in kinetic energy.
Wnet = ΔK = 23040 J
f) The speed of the rocket.
K = ½ mv2
23040 = ½ x 40 x v2
v2 = 23040/20
v = 33,9 m.s-1
speed = 33,9 m.s-1
40kg
80m
680 N
392 N
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10.
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Hi -
This is a SAMPLE presentation only.
My FULL presentations, which contain loads more slides and other resources, are freely
available on my resource sharing website:
www.sciencecafe.org.za
(paste into your browser if link above does not work)
Have a look and enjoy!
Keith Warne
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