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M.1.5WORK, ENERGY, and POWER

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WORK, ENERGY, and POWER
WORK
Work= Force x displacement.
W=F.d
Work = Product of displacement and force in the direction of
displacement.
W = (Fx ) (d)
W = (F cos Ο΄ )(d)
Work done on the block
a
F
π‘Šπ‘œπ‘Ÿπ‘˜ = 𝐹. 𝑑
d
𝐹
𝐹𝑦
Ο΄
a
π‘Šπ‘œπ‘Ÿπ‘˜ = (𝐹π‘₯ ) 𝑑
π‘Šπ‘œπ‘Ÿπ‘˜ = (𝐹 π‘π‘œπ‘ Ο΄) 𝑑
𝐹π‘₯
d
Work = F x d
1 π΅π‘‡π‘ˆ = 778 𝑓𝑑 − 𝑙𝑏
= 1.055 𝐾𝑗
= 0.252 πΎπ‘π‘Žπ‘™
Force, F= N, dyn, lb
Distance = m, cm, ft
Work = F x d
Work = N x m = N.m
= Joules (J)
= dyn x cm = (dyn-cm) or erg
Work = F x d
Work = lb x ft = lb-ft
or ft - lb
1 πΎπ‘π‘Žπ‘™ = 4.187𝐾𝑗
1 𝐾𝑗 = 1 𝐾𝑁 − π‘š
Example: Work done
A box is pushed without acceleration 5 m along a horizontal floor
against a frictional force of 180 N. How much work is done?
F=180 N
D=5 m
π‘Šπ‘œπ‘Ÿπ‘˜ = 𝐹. 𝑑
π‘Š = 180𝑁 π‘₯ 5π‘š = 900 𝑁 − π‘š
= 900 𝐽
= 0.9 𝐾 𝐽
Example: Work done
What work is performed in dragging a sled 50 ft horizontally without
acceleration when the force of 60 lb is transmitted by a rope making an
angle of 30 degrees with the ground.
π‘Šπ‘œπ‘Ÿπ‘˜ = 𝐹π‘₯ . 𝑑
F=60 lb
𝐹π‘₯
π‘Šπ‘œπ‘Ÿπ‘˜ = πΉπ‘π‘œπ‘ ∅ . 𝑑
30π‘œ
π‘Š = 60 𝑙𝑏 π‘π‘œπ‘ 30 (50 𝑓𝑑)
D=50 ft
π‘Š = 2598.076211 𝑓𝑑 − 𝑙𝑏
π‘“π‘œπ‘Ÿ π΅π‘‡π‘ˆ
1 π΅π‘‡π‘ˆ
π‘Š = 2598.076211 𝑓𝑑 − 𝑙𝑏 π‘₯
= 3.3394 π΅π‘‡π‘ˆ
778 𝑓𝑑 − 𝑙𝑏
ENERGY : The ability to do work
1. Potential Energy
- energy posses by virtue of its elevation.
Ex: electrical, elastic, chemical, and nuclear potential energy,
gravitational potential energy(the most common form)
2. Kinetic Energy
- energy posses by virtue of its velocity.
- is energy of motion.
POTENTIAL ENERGY, PE
Mass, m
Height, h
𝑃. 𝐸. = π‘Šβ„Ž
𝑃. 𝐸. = π‘šπ‘”β„Ž
𝑃. 𝐸. = π‘ƒπ‘œπ‘‘π‘’π‘›π‘‘π‘–π‘Žπ‘™ π‘’π‘›π‘’π‘Ÿπ‘”π‘¦, 𝐽
π‘š = π‘šπ‘Žπ‘ π‘ , 𝐾𝑔
β„Ž = β„Žπ‘’π‘–π‘”β„Žπ‘‘, π‘š
𝑔 = π‘Žπ‘π‘π‘’π‘™π‘’π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘› 𝑑𝑒𝑒 π‘‘π‘œ π‘”π‘Ÿπ‘Žπ‘£π‘–π‘‘π‘¦
9.81
π‘š
𝑓𝑑
π‘œπ‘Ÿ
32.2
𝑠2
𝑠2
π‘š
𝐾𝑔 − π‘š
𝑃. 𝐸. = 𝐾𝑔 2 π‘š =
π‘š =𝑁−π‘š =𝐽
2
𝑠
𝑠
Example: Potential Energy
• How much work is done in lifting a 5 Kg mass to a height of 30 m?
π‘Šπ‘œπ‘Ÿπ‘˜ π‘‘π‘œπ‘›π‘’ = 𝑃. 𝐸. = π‘Šβ„Ž
π‘Šπ‘œπ‘Ÿπ‘˜ π‘‘π‘œπ‘›π‘’ = 𝑃. 𝐸. = π‘šπ‘”β„Ž
π‘Šπ‘œπ‘Ÿπ‘˜ π‘‘π‘œπ‘›π‘’ = 5 π‘˜π‘” (9.81 π‘š 𝑠 2 )(30 π‘š)
= 1471.5
π‘˜π‘” − π‘š
(π‘š)
𝑠2
= 1471.5 𝑁 − π‘š
= 1471.5 𝐽
= 1.4715 𝐾𝐽
KINETIC ENERGY, K.E.
1
2
2
Δ𝐾. 𝐸. = π‘š 𝑣𝑓 − 𝑣𝑖
2
W
1
π‘Š
𝐾. 𝐸. = π‘šπ‘£ 2 =
𝑣2
2
2𝑔
v
m
𝐾. 𝐸. = 𝐾𝑖𝑛𝑒𝑑𝑖𝑐 π‘’π‘›π‘’π‘Ÿπ‘”π‘¦; 𝐽, 𝑓𝑑 − 𝑙𝑏
π‘š 𝑓
𝑣, 𝑣𝑓, 𝑣𝑖 = π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦, π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™, π‘“π‘–π‘›π‘Žπ‘™; ,
𝑠 𝑠
π‘š = π‘šπ‘Žπ‘ π‘ , π‘˜π‘”, 𝑠𝑙𝑒𝑔𝑠, π‘™π‘π‘š
1
𝐾. 𝐸. =
𝐾𝑔
2
π‘š
𝑠
π‘š
𝑓𝑑
𝑔 = π‘Žπ‘π‘π‘’π‘™π‘’π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘› 𝑑𝑒𝑒 π‘‘π‘œ π‘”π‘Ÿπ‘Žπ‘£π‘–π‘‘π‘¦; 9.81 2 , 32.2 2
𝑠
𝑠
2
π‘˜π‘” − π‘š2 π‘˜π‘” − π‘š
=
=
π‘š =𝑁−π‘š =𝐽
2
2
𝑠
𝑠
Example: KINETIC ENERGY
• What is the kinetic energy of a 800 Kg object which is moving at 27.78
m/s?
1
π‘Š 2
2
𝐾. 𝐸. = π‘šπ‘£ =
𝑣
2
2𝑔
𝐾. 𝐸. = 1 2 (800 𝐾𝑔) 27.78 π‘š 𝑠
𝐾. 𝐸. = 1 2 (800 𝐾𝑔) 27.78
𝐾. 𝐸. = 1 2 (800 𝐾𝑔) 27.78
2
2 π‘š2
2 π‘š2
2
𝐾. 𝐸. = 308691.36 𝐾𝑔 π‘š 𝑠 2
𝐾. 𝐸. = 308691.36 𝐽
𝑠2
𝑠2
𝐾. 𝐸. = 308.69136 𝐾𝐽
Example: KINETIC ENERGY
• Compute the kinetic energy of the 5 grams rifle traveling at 500 π‘š
1
π‘Š 2
2
𝐾. 𝐸. = π‘šπ‘£ =
𝑣
2
2𝑔
π‘šπ‘Žπ‘ π‘  = 5 𝑔 π‘₯
1 𝐾𝑔
= 0.005 π‘˜π‘”
1000𝑔
1
2
2
π‘š
𝐾. 𝐸. =
0.005 π‘˜π‘” (500)
𝑠2
2
𝐾. 𝐸. = 625
π‘˜π‘” − π‘š − π‘š
𝑠2
𝐾. 𝐸. = 625 𝑁 − π‘š π‘œπ‘Ÿ 𝐽
𝑠.
POWER
- the time rate of doing work.
π‘‘π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’
π‘ƒπ‘œπ‘€π‘’π‘Ÿ, 𝑃 = πΉπ‘œπ‘Ÿπ‘π‘’ π‘₯
π‘‘π‘–π‘šπ‘’
π‘š
𝑁−π‘š
𝐽
=𝑁
=
=
𝑠
𝑠
= π‘Šπ‘Žπ‘‘π‘‘π‘  (π‘Š)
π‘Šπ‘œπ‘Ÿπ‘˜
πΈπ‘›π‘’π‘Ÿπ‘”π‘¦
π‘ƒπ‘œπ‘€π‘’π‘Ÿ =
=
π‘‡π‘–π‘šπ‘’
π‘‘π‘–π‘šπ‘’
𝑠
1 𝐻𝑃 = 746 π‘€π‘Žπ‘‘π‘‘π‘ 
= 0.746 𝐾𝑀
𝑓𝑑−𝑙𝑏
= 33,000 π‘šπ‘–π‘›
=
𝐡𝑑𝑒
42.42 π‘šπ‘–π‘›
Example: POWER
A box is pushed across a horizontal floor by a horizontal force of 180 N.
• Find the power output in pushing the box 20 m in 5 minutes.
1 𝐻𝑃 = 746 π‘€π‘Žπ‘‘π‘‘π‘ 
= 0.746 𝐾𝑀
F=180 N
=
𝐽
= watts
𝑠
Distance=20 m
T=5 min.
π‘ƒπ‘œπ‘€π‘’π‘Ÿ =
π‘Šπ‘œπ‘Ÿπ‘˜
π‘‡π‘–π‘šπ‘’
𝐹. 𝑑
𝑑
180 𝑁 (20 π‘š)
π‘ƒπ‘œπ‘€π‘’π‘Ÿ =
300 𝑠𝑒𝑐
𝑓𝑑−𝑙𝑏
π‘šπ‘–π‘›
𝐡𝑑𝑒
42.42 π‘šπ‘–π‘›
= 33,000
𝑑 = 5 min π‘₯
60 𝑠𝑒𝑐
= 300 𝑠𝑒𝑐.
1 π‘šπ‘–π‘›
π‘ƒπ‘œπ‘€π‘’π‘Ÿ =
π‘ƒπ‘œπ‘€π‘’π‘Ÿ = 12
𝑁 − π‘š
𝑠𝑒𝑐
𝐽
= 12
𝑠𝑒𝑐
= 12 π‘Šπ‘Žπ‘‘π‘‘π‘ 
Example: POWER
A box is pushed across a horizontal floor by a force of 180 N making an angle of 30π‘œ with horizontal.
• Find the power output in pushing the box 20 m in 5 minutes.
F=180 N
1 𝐻𝑃 = 746 π‘€π‘Žπ‘‘π‘‘π‘ 
= 0.746 𝐾𝑀
30π‘œ
=
𝐽
= watts
𝑠
Distance=20 m
T=5 min.
π‘ƒπ‘œπ‘€π‘’π‘Ÿ =
π‘Šπ‘œπ‘Ÿπ‘˜
π‘‡π‘–π‘šπ‘’
𝑓𝑑−𝑙𝑏
π‘šπ‘–π‘›
𝐡𝑑𝑒
42.42 π‘šπ‘–π‘›
= 33,000
𝑑 = 5 min π‘₯
(πΉπ‘π‘œπ‘ θ)𝑑
𝑑
180 𝑁 π‘π‘œπ‘ 30 (20 π‘š)
π‘ƒπ‘œπ‘€π‘’π‘Ÿ =
300 𝑠𝑒𝑐
60 𝑠𝑒𝑐
= 300 𝑠𝑒𝑐.
1 π‘šπ‘–π‘›
π‘ƒπ‘œπ‘€π‘’π‘Ÿ =
= 10.39230485
𝑁 − π‘š
𝑠𝑒𝑐
= 10.39230485
𝐽
𝑠𝑒𝑐
= 10.39230485π‘Šπ‘Žπ‘‘π‘‘π‘ 
Efficiency, e
π‘ƒπ‘œπ‘€π‘’π‘Ÿ 𝑂𝑒𝑑𝑝𝑒𝑑 π‘ƒπ‘œ
𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦, 𝑒 =
=
π‘ƒπ‘œπ‘€π‘’π‘Ÿ 𝐼𝑛𝑝𝑒𝑑
𝑃𝑖
π‘Šπ‘œπ‘Ÿπ‘˜ 𝑂𝑒𝑑𝑝𝑒𝑑
π‘Šπ‘œπ‘Ÿπ‘˜ 𝐼𝑛𝑝𝑒𝑑
π‘šπ‘’π‘β„Žπ‘Žπ‘›π‘–π‘π‘Žπ‘™ 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦, ηπ‘š =
η𝑒
𝑃𝑖
η𝑒
π‘ƒπ‘œ
π‘šπ‘œπ‘‘π‘œπ‘Ÿ
𝑃𝑖
η𝐡
π‘π‘’π‘šπ‘
π‘ƒπ‘œ
𝑂𝑒𝑑𝑝𝑒𝑑
Example: Efficiency
• A power of 90 Kw is supplied to the motor of a crane. It lifts 6000 kg to a height of 15 m
in 18 sec. find the efficiency of the machine.
π‘ƒπ‘œπ‘€π‘’π‘Ÿ 𝑂𝑒𝑑𝑝𝑒𝑑 π‘ƒπ‘œ
𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦, 𝑒 =
=
π‘ƒπ‘œπ‘€π‘’π‘Ÿ 𝐼𝑛𝑝𝑒𝑑
𝑃𝑖
π‘ƒπ‘œ
𝑒=
𝑃𝑖
π‘š
πΈπ‘›π‘’π‘Ÿπ‘”π‘¦ 𝑃. 𝐸. π‘šπ‘”β„Ž 6000 π‘˜π‘” (9.81 𝑠 2 )(15 π‘š)
π‘ƒπ‘œ =
=
=
=
π‘‡π‘–π‘šπ‘’
𝑑
𝑑
18 𝑠
π‘ƒπ‘œ = 49050
𝐽
𝑠 π‘œπ‘Ÿ π‘€π‘Žπ‘‘π‘‘π‘ 
π‘ƒπ‘œ = 49.050 𝐾𝑀
𝑒=
π‘ƒπ‘œ 49.05 πΎπ‘Š
=
= 0.545 π‘œπ‘Ÿ 54.5%
𝑃𝑖
90 πΎπ‘Š
Example: Efficiency
• A power of 6 Kw is supplied to the motor of a crane. The motor has an efficiency
of 90%. With what constant speed does the crane lift 363 Kg weight?
π‘ƒπ‘œ
𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦, 𝑒 =
𝑃𝑖
π‘ƒπ‘œπ‘€π‘’π‘Ÿ, 𝑃 = πΉπ‘œπ‘Ÿπ‘π‘’ π‘₯
π‘‘π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’
= 𝐹(𝑣)
π‘‘π‘–π‘šπ‘’
π‘ƒπ‘œ
π‘ƒπ‘œ
𝑒 = = 0.90 =
𝑃𝑖
6 𝐾𝑀
π‘ƒπ‘œ = 5.4 π‘˜π‘€
π‘ƒπ‘œ = 𝐹(𝑣)
0.00981 𝐾𝑁
5.4 π‘˜π‘€ = (363 𝐾𝑔 π‘₯
)(𝑣)
1 π‘˜π‘”
𝐾𝑁 − π‘š
0.00981 𝐾𝑁
5.4
= (363 𝐾𝑔 π‘₯
)(𝑣)
𝑠
1 π‘˜π‘”
𝑣 = 1.516415194 π‘š 𝑠
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