U.41 PM P1 Address 430/2017 Lecture Question for No x cncrgy formu C Secure l ht
ID: 1616105 • Letter: U
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U.41 PM P1 Address 430/2017 Lecture Question for No x cncrgy formu C Secure l https// 5939557? module item id 507096 Apps Migrations of Green Comics by comixolo PLos Pathogens: Woo CALCULUS 11E G Waste Management a rashion Promotion M rukari's DeviantMrtce PS OBEY Seymour Stripe Zion & rt tan Lennox Home Online Discussion 12 m Account Section (Zoom) Announcements 16 m Dashboard Syllabus A block of mass m 1.36 kg slithers down an icy slope, as shown in the figure. The slant distance the block slides by is 20 m, during Assignments which it descends by a distance 12 m, and moves a horizontal distance (by Pythagoras' theorem) of 16 m. Resolving its weight into Courses components, the normal component of the gravitational force is N-0.8 m g and the tangential component is 0.6 m g. Grades How much does the potential energy decrease from the top to the bottom of the slope? People Groups 13.056 Piazza What is the frictional force on the block, if the coefficient of friction between the block and the ice is 0.403? Proctoru Portal Calendar What is the work done against the friction force by the time the block reaches the bottom of the slope? Inbox Ignoring all forces on the block apart from gravity and friction, what is its kinetic energy at the bottom of the slope? HelpExplanation / Answer
1.) Potential energy decreases form TOp to bottom = mgh
PE = 1.36*9.81*12 = 160.0992 J
2.)Frictional force = uN = 0.403*0.8mg = 5.37 N
3.)work done against friction = u*0.8mg*d = 0.403*0.8*1.36*9.81*20 = 86.026 J
4.)conserve energy
mgh = 0.5mv^2 + W
160.0992 = KE + 86.026
KE = 73.98 J
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