حلول الممارسات – 1 – الإستمرارية

عندما يتحرك الجسم لأسفل السطح المائل. فإن السطح المائل سيؤثر على الجسم بقوة احتكاك حركي، وليس بقوة احتكاك ساكن. 

سنرمز إلى زاوية ميل المستوى الذي يتحرك فيه الجسم بسرعة ثابتة بـ«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨22px¨»«mi mathvariant=¨bold¨ mathcolor=¨#0000FF¨»§#945;«/mi»«mo mathvariant=¨bold¨ mathcolor=¨#0000FF¨»`«/mo»«mo mathvariant=¨bold¨ mathcolor=¨#0000FF¨»`«/mo»«/mstyle»«/math» .

نستخدم هيئة محاور يكون فيها المحور X في اتجاه أسفل المستوى، والمحور Y في الاتجاه العمودي على المستوى. 

نحلل قوة الجاذبية لمركبتيها وفقًا لهيئة المحاور المُختارة. 

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نقسم المعادلتين ونعبر عن الزاوية «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨22px¨»«mi mathvariant=¨bold¨ mathcolor=¨#0000FF¨»§#945;«/mi»«mo mathvariant=¨bold¨ mathcolor=¨#0000FF¨»`«/mo»«mo mathvariant=¨bold¨ mathcolor=¨#0000FF¨»`«/mo»«/mstyle»«/math»:

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لذلك، فإن زاوية ميل المستوى الذي يتحرك به الجسم لأسفل المستوى بسرعة ثابتة هي 16.69 درجة.