Question

- BH Tài - TP 4th the ta! the slede an roberende Answer the following questi 4) Drast the tree body diagram for this scenario inding the best of the coordinates the direcaon of gravity with directlerie for the sled Label Alle ting on these including separated and comes Estimatex and y compose of the force vector, PE 5) Does the sled hare as acceleration is the directed s) Therefore, what will the sum of the forces in the directive equal [etle one) 0 4 Sum dhe forces in the direction and equestres question 38 .) Does the sled here as acceleration is the directa (side Therefore, what will the sum of the forces in the sequel (circle one) 0 2 Sum the forces is the direction and equate to the answer of the preus question to Use the equacions from and solve for e falling dont forget hi The normal force on the sled Answer

slope angle=20 degrees

friction coefficient= 0.1

mass pf sledriders=300 kg

Answer 1

F2 F3 C mo caso x component of gravity force Fore song sing = 300x9.8xsin 20 = 10o5.54N y component of gravity force Formgcaso =-300x 9.8 x cas 20 2762.7 N s means toward negative y acis mesino mg =- 62 yes sled have acceleration along x aces Sum of forces along a direction mg sino - f also friction force f=MFN No sled does not have acceleration along y curls Sum of fonce along y direction =

F2 (f) also friction force f= FN sled does not have acceleration along y cuels Sum of force along y direction = 0 OF F - mocaso =) F = ng caso a N (1) Normal force F = mg coso = 300x9.8x cos20 FN = 2762.7 N friction force fe FN =-0. 1x2762.7 =-276-27 N along negative ac direction mesino - f Z 300 a = 300x9.8× Sin2o- 276.27 a = 2.43 m/s² slope is increased friction force decreases and mgsind increases so acceleration increases Family will be thrown forward. This is First on example of Neuton third low. Each action has equal and opt

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