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A detailed explanation and calculations related to the work-energy theorem and potential energy in physics, including various examples and formulas for different types of forces and movements. It covers topics such as work done by a constant force, work done by a variable force, work done by gravity, and work done by friction, among others.
Typology: Exercises
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واﺳﺘﻨﺘﺞ ﻃﺒﯿﻌﺔ اﻟﺸﻐﻞ ﻓﻲ ﻛﻞAB=25cmﺧﻼل اﻹﻧﺘﻘﺎلF=10Nﺷﺪﺗﮫﺎ
اﻟﻰ ﻧﻘﻄﺔA ﺧﻼل إﻧﺘﻘﺎل ﻣﺮﻛﺰ ﻗﺼﻮه ﻣﻦ ﻧﻘﻄﺔm=10gﻛﺘﻠﺘﻪ(S) أﺣﺴﺐ ﺷﻐﻞ وزن ﺟﺴﻢ ﺻﻠﺐ
g=10N.kg ﺷﺪة اﻟﺜﻘﺎﻟﺔ: ﻧﻌﻄﻲ
ﻓﺘﻜﻮن ﺣﺮﻛﺔ ﻣﺮﻛﺰ ﻗﺼﻮرھﺎ ﺣﺴﺐm=8,5kg ﻳﻘﻮم رﻳﺎﺿﻲ أﺛﻨﺎء ﺗﺪارﻳﺒﻪ ﺑﺮﻣﻲ ﻛﺮة ﺣﺪﻳﺪﻳﺔ ﻛﺘﻠﺘﮫﺎ
y :ﺗﻐﺎدر اﻟﻜﺮة ﻳﺪ اﻟﺮﻳﺎﺿﻲ أرﺗﻮﺑﮫﺎO اﻟﻨﻘﻄﺔ
O
. =1,90m
(y :ھﻲ أﻋﻠﻰ ﻧﻘﻄﺔ ﻓﻲ اﻟﻤﺴﺎر إﺣﺪاﺛﯿﺎﺗﮫﺎ S
S
=4,5m;x
S
. =6,72m)
x :ﻣﺪى اﻟﺤﺮﻛﺔ أﻓﺼﻮھﺎ D
D
. =16,20m
→
ﻪﺳﻤﻜm=10kgوﻛﺘﻠﺘﻪL=4m ﻧﻌﺘﺒﺮ ﺳﻠﻤﺎ ﻃﻮﻟﻪ
g=9,81N.kg ﻧﺄﺧﺬ
ﻗﺎﺑﻠﺔ ﻟﻠﺪورانL=50,0cmوﻃﻮﻟﮫﺎm=200g ﻧﻌﺘﺒﺮ ﺳﺎﻗﺎ ﻣﺘﺠﺎﻧﺴﺔ ﻛﺘﻠﺘﮫﺎ
ﻮنﻧﺤﺮر اﻟﺴﺎق ﻣﻦ ﻣﻮﺿﻊ ﻳﻜ.Oxﺣﻮل اﻟﻤﺤﻮر اﻷﻓﻘﻲ ﺑﺪزن اﺣﺘﻜﺎك
زن اﻟﺴﺎق ﺑﯿﻦوأﺣﺴﺐ ﺷﻐﻞOzﻣﻊ اﻟﻤﺤﻮر اﻟﺮأﺳﻲ = 45 ° زاوﻳﺔ
.Ozواﻟﻮﺿﻌﯿﺔ اﻟﺘﻲ ﻳﺘﻄﺎﺑﻖ اﺗﺠﺎھﮫﺎ ﻣﻊ اﺗﺠﺎه اﻟﻤﺤﻮر ھﺬه اﻟﻮﺿﻌﯿﺔ
g=9.81N.kg ﻧﺄﺧﺬ
،ﻣﻌﻠﻖ ﺑﺨﯿﻂ ﻛﺘﻠﺘﻪ ﻣﮫﻤﻠﺔ وﻏﯿﺮ ﻗﺎﺑﻞ ﻟﻠﻤﺪm=50g ﺟﺴﻢ ﻧﻘﻄﻲ ﻛﺘﻠﺘﻪ
.L=40cm ﻃﻮﻟﻪ
v=0,5m.s :ﻗﺪرة اﻟﻤﺤﺮك ، ﻋﻠﻤﺎ أن ﺳﺮﻋﺔ اﻟﺤﻤﻮﻟﺔ ھﻲأﺳﺘﻨﺘﺞ
1000 ﺑﺴﺮﻋﺔ ﺛﺎﺑﺘﺔ ﺗﺴﺎويD=10cmﻧﺪﻳﺮ ﻗﺮﺻﺎ ﻣﺘﺠﺎﻧﺴﺎ ﻗﻄﺮهP=1kW ﺑﻮاﺳﻄﺔ ﻣﺤﺮك ﻗﺪرﺗﻪ
.أﺳﺘﻨﺘﺞ اﻟﺴﺮﻋﺔ اﻟﺰاوﻳﺔ ﻟﻠﻘﺮص.Hzﻟﺪوران اﻟﻘﺮص ﺑﺎﻟﻮﺣﺪةNأﺣﺴﺐ اﻟﺘﺮدد - 1
W(𝐹⃗) = 𝐹. 𝐴𝐵. cos(𝐹⃗ , 𝐴𝐵
−
× cos 60° = 1,25𝐽
−
× cos 90° = 0
−
× cos(180° − 45°) = −1,75𝐽
−
× cos(90° − 30°) = 1,25𝐽
−
× cos 0° = 2,5𝐽
−
× cos 180° = −2,5𝐽
𝐴
𝐵
A→ 𝐵
𝑂
𝑆
O→ 𝑆
O→ 𝑆
𝑂
𝐷
O→ 𝐷
𝑜
𝑀
O→ 𝑀
𝑂→𝑀
𝑂
𝑀
𝑂
𝑀
𝑂→𝑀
𝑂
𝑀
𝑂
𝑀
1
1
الى الموضعz 2
2
.z
1
2
cos𝛼 =
ℎ
𝐿
2
𝐿
2
cos 𝛼
𝐿
2
cos 𝛼
4
2
cos 30° = 169,7𝐽
1
2
1
2
cos 𝛼 = مع :
𝑧
1
𝐿
2
𝐿
2
= cos 𝛼 1
− و z
𝐿
2
2
z
𝐿
2
cos 𝛼 − (−
𝐿
2
𝐿
2
(cos 𝛼 + 1)
−
−
2
(cos 45° + 1)
𝐴→𝐵
𝐴
𝐵
𝐴
=OC-OH=L-L.cos𝛼
𝐴
𝐵
𝐴
𝐵
sin 𝛼 =
ℎ
𝐴𝐵
ℎ = 𝐴𝐵 sin 𝛼 ومنه :
𝐴
𝐵
= −ℎ = −𝐴𝐵 sin 𝛼
𝐴→𝐵
) = −𝑚𝑔𝐴𝐵 sin 𝛼
𝐴→𝐵
) = −80 × 10 × 1500 × sin 20° = −4,1. 10
4
𝐴→𝐵
𝑊
𝐴→𝐵
𝑊
= 𝑓. 𝐴𝐵. cos 𝜋
𝐴→𝐵
𝑊
𝐴→𝐵
(𝑓
⃗ )
4
𝑁
𝑁
𝑁
4
4
4
𝑁
𝑁
نحصل على :xx’ بإسقاط العالقة المتجهية على المحور
−𝑃 sin 𝛼 − 𝑓 + 𝑇 = 0
𝑇 = 𝑃 sin 𝛼 + 𝑓
𝑇 = 1000 sin 30° + 200 = 700 𝑁 ت.ع:
𝑚
∆
′
⃗⃗⃗⃗
) + 𝑀
∆
′
⃗⃗⃗⃗
) + 𝑀
∆
′
⃗⃗⃗⃗⃗
) + 𝑀
𝑚
∆
′
⃗⃗⃗⃗
∆
′ ⃗⃗⃗⃗
𝑚
𝑚
∆
∆
𝐷
2
