A particle moves in the xy plane as shown in the figure below and experiences a

Question

                 A particle moves in the xy plane as shown in the figure below and experiences a friction force with constant magnitude 6.87 N, always in the direction opposite the object's velocity. Calculate the work that you must do to slide the object at constant speed against the friction force as the object moves along the following paths.

A particle moves in the xy plane as shown in the figure below and experiences a friction force with constant magnitude 6.87 N, always in the direction opposite the object's velocity. Calculate the work that you must do to slide the object at constant speed against the friction force as the object moves along the following paths. the purple path O to followed by a return purple path to O the purple path O to followed by a return blue path to O the blue path O to followed by a return blue path to O

Explanation / Answer

Total Force = mass*acceleration Total Force =mass*0 Total Force = 0 Frictional force = 6.87 N Total Force = Force - Frictional Force = 0 so Force = Frictional Force Force = 6.87 N To find work we use the following formula: Work = Force * distance Part A: Using the coordinates given, we realize that the distance from O to A is 5 m, since the object goes from O to A and then back to O, we multiply 5 by 2, and get 10 m for the total distance (net displacement) traveled in part A. distance = 10 m Work = 6.87 N * 10m = 68.7 J Part B. First find the net displacement (total distance traveled.) The purple path from O to C is 10 m. And the return blue path can be found using the pythageoreom theorem: a^2+b^2=c^2 ?(25m+25m) = c c=7m Therefore the net displacement is 5m + 5m + 7m = 17m Plugging this into the work equation we get: Work = 6.87 N* 17m = 116.79 J Part C is also solved the same way. Work =   6.87 N * 14 m = 96.18 J To find work we use the following formula: Work = Force * distance Part A: Using the coordinates given, we realize that the distance from O to A is 5 m, since the object goes from O to A and then back to O, we multiply 5 by 2, and get 10 m for the total distance (net displacement) traveled in part A. distance = 10 m Work = 6.87 N * 10m = 68.7 J Part B. First find the net displacement (total distance traveled.) The purple path from O to C is 10 m. And the return blue path can be found using the pythageoreom theorem: a^2+b^2=c^2 ?(25m+25m) = c c=7m Therefore the net displacement is 5m + 5m + 7m = 17m Plugging this into the work equation we get: Work = 6.87 N* 17m = 116.79 J Part C is also solved the same way. Work =   6.87 N * 14 m = 96.18 J

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