The words you are searching are inside this book. To get more targeted content, please make full-text search by clicking here.
Discover the best professional documents and content resources in AnyFlip Document Base.
Search
Published by sgillet.iscles, 2024-01-08 09:05:11

T_spe_C09_BE2_cor

T_spe_C09_BE2_cor

T spe – PC – Chapitre 9 – banque d’exercices Page 3 Corrigé Exercice 1 ① On considère le système {balle} dans le référentiel terrestre, supposé galiléen. ② On choisit 2 moments clé pour la balle : • point A : zA = 1,50 m et vA = ? (point de départ) • point B : zB = 5,0 m et vB = 0 car c’est le point le plus haut ③ On exprime l’énergie mécanique à ces deux moments clés : • point A : = + = 1 2 ⋅ ⋅ 2 + ⋅ ⋅ point B : = + = 1 2 ⋅ ⋅ 2 + ⋅ ⋅ = ⋅ ⋅ car vB = 0 ④ On néglige les frottements : l’énergie mécanique se conserve. On peut donc écrire : = 1 2 ⋅ ⋅ 2 + ⋅ ⋅ = ⋅ ⋅ 1 2 ⋅ 2 + ⋅ = ⋅ = �2 ⋅ (− ) AN : = �2 × 9,81 ⋅ (5,0 − 1,50) soit = 8,3. −1 Exercice 2 1. Expression de la variation d’énergie mécanique pendant le freinage : 2. Calcul de la valeur de la force de freinage f : = ⋅ 2 2 ⋅ = 1000 × � 83,5 3,6 � 2 2 × 50,0 = 5.4. 103 Exercice 3 Δ Em=E mB −EmA =1 2 ⋅m⋅vB 2 +m⋅g⋅zB−1 2 ⋅m⋅v A 2 −m⋅g⋅zA =EcB +E ppB −(EcA +E pp A ) =1 2 ⋅m×0+m⋅g⋅z A−1 2 ⋅m⋅v 2 −m⋅g⋅z A Δ Em=−1 2 ⋅m⋅v2 −1 2 ⋅m⋅v2 =⃗f⋅AB⃗ Δ Em=W AB (⃗f) −1 2 ⋅m⋅v2 = f⋅AB⋅cos(180) −1 2 ⋅m⋅v 2 =− f⋅AB 1 2 ⋅m⋅v 2 = f⋅AB


T spe – PC – Chapitre 9 – banque d’exercices Page 4 Exercice 4 Exercice 5 Exercice 6 Exercice 7


Click to View FlipBook Version