[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"detail-sidebar-cat-0-en-105":3,"doc-seo-312046-105":59,"doc-detail-312046-en":131},{"code":4,"msg":5,"data":6},0,"success",[7,13,18,23,28,33,38,43,48,51,55],{"id":8,"doc_module":4,"doc_module_name":9,"category_name":10,"show_sort_weight":11,"slug":12},1,"Document","Story & Novel",90,"story-novel",{"id":14,"doc_module":4,"doc_module_name":9,"category_name":15,"show_sort_weight":16,"slug":17},2,"Literature",80,"literature",{"id":19,"doc_module":4,"doc_module_name":9,"category_name":20,"show_sort_weight":21,"slug":22},4,"Exam",70,"exam",{"id":24,"doc_module":4,"doc_module_name":9,"category_name":25,"show_sort_weight":26,"slug":27},5,"Comic",60,"comic",{"id":29,"doc_module":4,"doc_module_name":9,"category_name":30,"show_sort_weight":31,"slug":32},6,"Technology",50,"technology",{"id":34,"doc_module":4,"doc_module_name":9,"category_name":35,"show_sort_weight":36,"slug":37},7,"Healthcare",40,"healthcare",{"id":39,"doc_module":4,"doc_module_name":9,"category_name":40,"show_sort_weight":41,"slug":42},8,"Research & Report",30,"research-report",{"id":44,"doc_module":4,"doc_module_name":9,"category_name":45,"show_sort_weight":46,"slug":47},9,"Religion & Spirituality",20,"religion-spirituality",{"id":46,"doc_module":4,"doc_module_name":9,"category_name":49,"show_sort_weight":46,"slug":50},"World Cup","world-cup",{"id":52,"doc_module":4,"doc_module_name":9,"category_name":53,"show_sort_weight":52,"slug":54},10,"Lifestyle","lifestyle",{"id":56,"doc_module":4,"doc_module_name":9,"category_name":57,"show_sort_weight":24,"slug":58},19,"General","general",{"code":4,"msg":60,"data":61},"ok",{"site_id":62,"language":63,"slug":64,"title":65,"keywords":66,"description":67,"schema_data":68,"social_meta":124,"head_meta":126,"extra_data":128,"updated_unix":130},105,"en","flipping-physics-lecture-notes-mechanics-free-response-question-3-solutions-ap-physics-c-1998-released-exam","Flipping Physics Lecture Notes - Mechanics Free Response Question 3 Solutions - AP Physics C 1998 Released Exam","","Mechanics free response question solutions for AP Physics C 1998, presenting a step-by-step Newton’s laws approach. The walkthrough defines variables and positive directions, constructs complete free body diagrams, and derives normal forces, tensions, and friction conditions for both static and kinetic cases. It computes relations like N1 = m1g, N2 = (m1 + m2)g, and uses friction constraints (µs2, µk1, µk2) to solve for acceleration and determine how behavior changes when blocks begin slipping relative to each other.",{"@graph":69,"@context":123},[70,84,106],{"@type":71,"itemListElement":72},"BreadcrumbList",[73,77,79,82],{"item":74,"name":75,"@type":76,"position":8},"https://docshare.wps.com","Home","ListItem",{"item":78,"name":9,"@type":76,"position":14},"https://docshare.wps.com/document/",{"item":80,"name":20,"@type":76,"position":81},"https://docshare.wps.com/document/exam/",3,{"item":83,"name":65,"@type":76,"position":19},"https://docshare.wps.com/document/flipping-physics-lecture-notes-mechanics-free-response-question-3-solutions-ap-physics-c-1998-released-exam/312046/",{"url":83,"name":65,"@type":85,"image":86,"author":91,"headline":65,"publisher":94,"fileFormat":97,"inLanguage":63,"description":67,"dateModified":98,"datePublished":99,"encodingFormat":97,"isAccessibleForFree":100,"interactionStatistic":101},"DigitalDocument",{"url":87,"@type":88,"width":89,"height":90},"https://docshare.wps.com/thumbnails/flipping-physics-lecture-notes-mechanics-free-response-question-3-solutions-ap-physics-c-1998-released-exam/312046.png","ImageObject",300,407,{"name":92,"@type":93},"Bill Black","Person",{"url":74,"name":95,"@type":96},"DocShare","Organization","application/pdf","2026-11-04","2026-09-20",true,{"@type":102,"interactionType":103,"userInteractionCount":105},"InteractionCounter",{"@type":104},"ViewAction",11,{"@type":107,"mainEntity":108},"FAQPage",[109,115,119],{"name":110,"@type":111,"acceptedAnswer":112},"How are the normal forces N1 and N2 determined in part (a)?","Question",{"text":113,"@type":114},"For block 1, applying ∑Fy = 0 gives N1 = m1g. For block 2, applying ∑Fy = 0 yields N2 = m2g + N1 = (m1 + m2)g.","Answer",{"name":116,"@type":111,"acceptedAnswer":117},"What friction relationships are used to connect µs2 with the masses in the static case?",{"text":118,"@type":114},"Because the blocks are not slipping relative to one another, friction on block 2 is static and balances other forces, leading to µs2 N2 = mhg. Substituting N2 = (m1 + m2)g gives mh = µs2 (m1 + m2).",{"name":120,"@type":111,"acceptedAnswer":121},"How does the solution change when the blocks start slipping relative to each other?",{"text":122,"@type":114},"Friction forces switch from static to kinetic: Ff1 becomes kinetic with µk1 and Ff2 becomes kinetic with µk2. The analysis recalculates accelerations by applying Newton’s 2nd law separately (e.g., a1x = µk1g) and then combining forces on the combined m2 and hanging mass to solve for the system acceleration.","https://schema.org",{"og:url":83,"og:type":125,"og:title":65,"og:site_name":95,"og:description":67},"article",{"robots":127,"canonical":83},"index,follow",{"doc_id":129,"site_id":62},312046,1790698814,{"code":4,"msg":5,"data":132},{"doc_id":129,"user_id":133,"nickname":92,"user_avatar":134,"doc_module":4,"category_id":19,"category_name":20,"doc_title":65,"doc_description":67,"doc_content":135,"file_id":136,"file_url":137,"file_type":138,"file_size":139,"view_count":105,"is_deleted":4,"is_public":8,"is_downloadable":8,"audit_status":8,"page_count":14,"language":140,"language_code":63,"site_id":62,"html_lang":63,"table_of_contents":141,"faqs":142,"seo_title":143,"seo_description":67,"update_tm":144,"read_time":24,"preview_page_count":4},24189269381491,"https://ap-avatar.wpscdn.com/avatar/160000cf11732dd8392?x-image-process=image/resize,m_fixed,w_180,h_180&k=1788146458752108895","Flipping Physics Lecture Notes: Mechanics Free Response Question \\#3 Solutions AP Physics C 1998 Released Exam from the College Board  \nAP® is a registered trademark of the College Board , which was not involved in the production of, and does not endorse, this product.  \nFor legibility, please observe this legend: f1 = Ff1 , f2 = Ff2 , & M = mh ( for mass hanging)  \nKnowns, m 1 , m2 , mh, µs1 , µk1 , µs2 , µk2 . Give answers in terms of these variables and g.  \nComplete Free Body Diagram:  \n(a) v = 0  \n(a i) The normal force N 1 exerted on block 1 by block 2 is up and acts on the bottom surface of m 1 .  \n∑ Fy (on m1) = N1 − Fg1 = m1a1y = m1 (0 ) = 0 ⇒ N1 = Fg1 = m1g  \n(a ii) Because m 1 is not moving, there is no friction force Ff1 exerted on block 1 by block 2. Ff1 = 0  \n(a iii) The Tension force T exerted by a string is always a pull in the direction of the string. Therefore T is to the right and acts right where the string attaches to block 2. Let’s set to the right on m2 and down on mh as the positive direction.  \n∑ Fy (on mh ) = −T + Fgh = mhay = mh (0 ) = 0 ⇒ T = Fgh = mhg (a iv) The normal force N2 exerted on block 2 by the tabletop is up and acts on the bottom surface of m2 .  \n∑ Fy (on m2 ) = N2 − Fg 2 − N1 = m2 a2 y = m2 (0 ) = 0  \n⇒ N2 = Fg 2 + N1 = m2 g + m1g ⇒ N2 = (m2 + m1 ) g  \n(a v) The friction force Ff2 exerted on block 2 by the tabletop is keeping the block from moving and is therefore to the left and acts on the bottom surface of m2.  \n∑ Fx (on m2 ) = T − Fsf2 = m2 a2 x = m2 (0 ) = 0 ⇒ Fsf2 = T = mhg  \n(b) Sum the forces on all the objects in the positive direction indicated on the pulley in the diagram.  \n∑ Fall in +dir = −Fsf2 + T − T + Fgh = mta = mt (0 ) = 0 ⇒ Fsf2 = Fgh  \n⇒ µs2N2 = mhg ⇒ µs2 (m1g + m2 g ) = mhg ⇒  \nmh = µs2 (m1 + m2 )  \n0042 Lecture Notes-Mechanics FRQ \\#3 Solutions-AP Physics C 1998 Released Exam.docx page 1 of 2  \n(c) The blocks are now accelerating, however, they are not slipping relative to one another and therefore have the same acceleration in the x direction. The only change in the Free Body Diagram is the addition of two friction forces Ff1 between block 1 and block 2 (because they area Newton’s 3rd Law interaction pair) . Ff1 on block 2 is to the left and acts on the top of block 2. Ff1 on block 1 is to the right and acts on the bottom of block 1.  \nBecause there are no additional forces in the y-direction on either block, both Newton’s 2nd law equations in the y  \ndirection are still valid. Therefore N1 = m1g and N2 = m1g + m2 g .  \nNow we sum the forces on the whole system in the same direction we had defined as positive before , which is to the right on block 2 and down on the mass hanging.  \n∑ Fx system = Fsf1 − Fsf1 + T − Fkf2 −T + FgM = mtotalatotal ⇒ −µk 2N2 + mhg = (m1 + m2 + mh )a  \n− µk 2 (m1g + m2 g ) + mhg ⇒ a = ~~ ~~ =  \nm1 + m2 + mh  \nmh~~ ~~−m1~~ ~~k~~ ~~2m(1~~ ~~+~~ ~~mm~~ ~~h2~~ ~~) ( g )  \n(d) Now the two blocks are slipping relative to one another; this changes Ff1 to kinetic friction. Again there are no additional forces in the y-direction on either block, both Newton’s 2nd law equations in the y-direction are still valid. Therefore N1 = m1g and N2 = m1g + m2 g .  \n(d i) ∑ Fx m1 = Fkf1 = m1a1x ⇒ µk1N1 = µk1m1g = m1a1x ⇒ a1x = µk1g  \n(d ii) Now we sum the forces on m2 and mh again in the“positive direction”.  \n∑ Fm2 & M +dir = −Fkf1 − Fkf2 + T − T + Fgh = (m2 + mh )am2 & mh ⇒ −µk1N1 − µk 2N2 + mhg = (m2 + mh )a  \n⇒ −µk1m1g − µk 2 (m1g + m2 g ) + mhg = (m2 + mh )a ⇒ a =  \n( g ) (mh − µk1m1 − µk 2 (m1 + m2 ))(m2 + mh )  \nNotes about the published solutions:  \n(a) Be careful to indicate where the forces act on the objects. Students often forget and have them all acting on the center of mass of the object.  \n-This whole problem is really just drawing Free Body Diagrams and using Newton’s 2nd law repeatedly. This should indicate to you the importance of those very basic skills.","cbCain4CnHiydQP8","https://ap.wps.com/l/cbCain4CnHiydQP8","pdf",2718058,"English","# Known Variables and Free Body Diagram Setup\n# Static Case: No Slipping Relative to One Another\n# System Acceleration Derivation\n# Kinetic Case: Blocks Slip Relative to Each Other\n# Notes on Solution Approach","[{\"question\":\"How are the normal forces N1 and N2 determined in part (a)?\",\"answer\":\"For block 1, applying ∑Fy = 0 gives N1 = m1g. For block 2, applying ∑Fy = 0 yields N2 = m2g + N1 = (m1 + m2)g.\"},{\"question\":\"What friction relationships are used to connect µs2 with the masses in the static case?\",\"answer\":\"Because the blocks are not slipping relative to one another, friction on block 2 is static and balances other forces, leading to µs2 N2 = mhg. Substituting N2 = (m1 + m2)g gives mh = µs2 (m1 + m2).\"},{\"question\":\"How does the solution change when the blocks start slipping relative to each other?\",\"answer\":\"Friction forces switch from static to kinetic: Ff1 becomes kinetic with µk1 and Ff2 becomes kinetic with µk2. The analysis recalculates accelerations by applying Newton’s 2nd law separately (e.g., a1x = µk1g) and then combining forces on the combined m2 and hanging mass to solve for the system acceleration.\"}]","Flipping Physics Lecture Notes - Mechanics Free Response Question 3 Solutions - AP Physics C 1998 Released Exam | PDF",1789895153]