# Sarah

619 mots 3 pages
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‫1( ﻧﺺ ﻣﺒﺮهﻨﺔ اﻟﻄﺎﻗﺔ اﻟﺤﺮآﻴﺔ:‬ ‫ﺗﻐﻴﺮ اﻟﻄﺎﻗﺔ اﻟﺤﺮآﻴﺔ ﻟﺠﺴﻢ ﺻﻠﺐ ﻓﻲ إزاﺣﺔ أو ﻓﻲ دوران ﺣﻮل ﻣﺤﻮر ﺛﺎﺑﺖ، ﺑﻴﻦ ﻟﺤﻈﺘﻴﻦ ﻳﺴﺎوي اﻟﻤﺠﻤﻮع اﻟﺠﺒﺮي ﻷﺷﻐﺎل‬ ‫آﻞ اﻟﻘﻮى اﻟﻤﻄﺒﻘﺔ ﻋﻠﻰ اﻟﺠﺴﻢ ﺑﻴﻦ هﺎﺗﻴﻦ اﻟﻠﺤﻈﺘﻴﻦ:‬ ‫‪ΔEC = EC ( f ) − EC (i ) = ∑ W‬‬ ‫‪i→ f‬‬ ‫2( ﺷﻐﻞ وزن )‪ (S‬ﺑﻴﻦ 1 ‪ M‬و 4 ‪: M‬‬
‫4 ‪M1→ M‬‬

‫) ‪W ( P) = m g ( Z − Z‬‬ ‫1‬ ‫4‬ ‫ﺣﺴﺐ اﻟﺸﻜﻞ اﻟﺘﺎﻟﻲ:‬

‫ﻧﺠﺪ :‬ ‫* ‪Z1 − Z 4 = h‬‬ ‫‪h‬‬ ‫‪= sin α‬‬ ‫4 ‪M 1M‬‬ ‫وﻣﻨﻪ ﻧﻜﺘﺐ : ‪Z1 − Z 4 = M 1M 4 .sin α‬‬
‫4 ‪M1→ M‬‬ ‫4 ‪M1→ M‬‬

‫إذن: ∝ ‪W ( P) = m g .M 1M 4 .sin‬‬

‫ت.ع : °42 ‪W ( P) = 0, 4 ×10 × 5, 4.10−2 × sin‬‬

‫‪W ( P ) ≈ 8,8.10 −2 J‬‬ ‫4 ‪M1 → M‬‬

‫3( ﻟﻨﺠﺪ ﺳﺮﻋﺔ اﻟﺠﺴﻢ )‪ (S‬ﻋﻨﺪ آﻞ ﻣﻦ اﻟﻤﻮﺿﻌﻴﻦ 1 ‪ M‬و 4 ‪. M‬‬ ‫‪M M‬‬ ‫‪M M‬‬ ‫و 5 3 = 4‪v‬‬ ‫2 0 = 1‪v‬‬ ‫‪2τ‬‬ ‫‪2τ‬‬ ‫2−‬ ‫01.8,4‬ ‫2−01.4 ,2‬ ‫= 4‪v‬‬ ‫= 1‪v‬‬ ‫و‬ ‫40 ,0 × 2‬ ‫40 ,0 × 2‬ ‫1−‬ ‫1− ‪ v1 = 0,3m.s‬و‬ ‫‪v4 = 0, 6m.s‬‬ ‫إذن اﻟﻄﺎﻗﺔ اﻟﺤﺮآﻴﺔ ل )‪ (S‬ﻋﻨﺪ آﻞ ﻣﻦ اﻟﻤﻮﺿﻌﻴﻦ 1 ‪ M‬و 4 ‪ M‬هﻲ ﻋﻠﻰ اﻟﺘﻮاﻟﻲ:‬

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‫‪cours pratiques en ligne‬‬
‫= ‪EC‬‬
‫4‬

‫2 1‬ ‫و 4‪mv‬‬ ‫2‬

‫= ‪EC‬‬
‫1‬

‫2 1‬ ‫1‪mv‬‬ ‫2‬

‫1‬ ‫1‬ ‫2 )3,0( × 4 ,0 × = 1‪ EC‬و 2 )6 ,0( × 4 ,0 × = 4 ‪EC‬‬ ‫2‬ ‫2‬ ‫2−‬ ‫و ‪EC = 7, 2.10 J‬‬ ‫‪EC = 1,8.10− 2 J‬‬
‫4‬ ‫1‬

‫4( إﻳﺠﺎد ﺷﻐﻞ اﻟﻘﻮة ‪ R‬اﻟﻤﻄﺒﻘﺔ ﻣﻦ ﻃﺮف اﻟﻤﻨﻀﺪة ﻋﻠﻰ اﻟﺠﺴﻢ )‪: (S‬‬ ‫ﻧﻄﺒﻖ ﻣﺒﺮهﻨﺔ اﻟﻄﺎﻗﺔ اﻟﺤﺮآﻴﺔ ﻋﻠﻰ )‪.(S‬‬ ‫ﻧﺄﺧﺬ:‬ ‫ اﻟﺤﺎﻟﺔ اﻟﺒﺪﺋﻴﺔ )‪ : (i‬وﺟﻮد )‪ (S‬ب 1 ‪M‬‬‫ اﻟﺤﺎﻟﺔ اﻟﻨﻬﺎﺋﻴﺔ )‪ : (f‬وﺟﻮد )‪ (S‬ب 4 ‪M‬‬‫اﻟﺠﺴﻢ )‪ (S‬ﻳﺨﻀﻊ ﺧﻼل ﺣﺮآﺘﻪ ﻟﻠﻘﻮﺗﻴﻦ ‪ p‬و ‪ R‬ﻓﻘﻂ:‬ ‫إذن ﺳﻨﻜﺘﺐ ﺻﻴﻐﺔ اﻟﻤﺒﺮهﻨﺔ ﺑﻴﻦ 1 ‪ M‬و 4 ‪: M‬‬

‫)‪EC − Ec = W ( P) + W ( R‬‬ ‫1‬ ‫4‬
‫‪W R = E −E − W P‬‬ ‫‪C‬‬ ‫‪C‬‬ ‫4‬ ‫4 ‪1 M1→M‬‬

‫4 ‪M1→M‬‬

‫) (‬

‫4 ‪M1 → M‬‬

‫4 ‪M1 → M‬‬

‫) (‬

‫ت.ع : 2−01.8,8 − 2−01.8,1 − 2− 01.2 ,7 = ) ‪W ( R‬‬ ‫4 ‪M1 → M‬‬ ‫‪W ( R ) = −3, 4.10−2 J‬‬ ‫4 ‪M1 → M‬‬

‫5( ﻧﺴﺘﻨﺘﺞ أن ﻃﺒﻴﻌﺔ اﻟﺘﻤﺎس ﺑﻴﻦ اﻟﺠﺴﻢ )‪ (S‬واﻟﻤﻨﻀﺪة ﻳﺘﻢ ﺑﺎﻻﺣﺘﻜﺎك ﻷن ﺷﻐﻞ اﻟﻘﻮة ‪ R‬ﺳﺎﻟﺐ.‬ ‫6( ﻃﺎﻗﺔ اﻟﻮﺿﻊ اﻟﺜﻘﺎﻟﻴﺔ ﻟﻠﺠﺴﻢ )‪ (S‬ﻓﻲ اﻟﻤﻮﺿﻊ ‪ M‬ذي اﻷﻧﺴﻮب ‪: Z‬‬ ‫ﻧﻌﺒﺮ ﻋﻨﻬﺎ ﺑﺎﻟﻌﻼﻗﺔ اﻟﺘﺎﻟﻴﺔ :‬ ‫‪EP = m

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