Constant Horizontal Force - General Physics - Solved Exam, Exams of Physics

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Q.1. A box of M = 40-kg mass is pulled for a distance of s = 5.0 m along a horizontal floor with a constant horizontal force of F = 140 N. The kinetic coefficient of friction between the box and the floor is pi, = 0.30. a) Calculate the work done by the applied force. We = Fd coso = (iho w) (Sm) lt) = 00 Fouler We = F00F ; ae ee b) Calculate the work done by the friction force. a) Wy = {d ees! B0" = (Hy mq) a €23130" We = (03%40x10 N) (Sm) (-1) = — 600 foules We =— S007 c) When the force of F = 140 N is applied initially, the box was moving at v,= 2.0 m/s in the same direction with the force. Calculate the final speed of the box. . I zu Waet = Ke -Ki = z Moy? ~ eM, se 2 $00 —600 = 20Ux"- 80 Sy=3 = Uy = 3 w/s Up = 2 m/s / d) Calculate the instantaneous power delivered by the force F, at the moment when it is applied initially => FS P =F =(jhon) (2m) e230 = 280 warts p= 250w e) Calculate the average power delivered by the applied force F, for the distance of s = 5.0 m Ue=v. tat — whee gx Fad = o-i2e atm \ . mM Ao a ¢ So, 3% ay +yt & ta2 Seconds Hence, the avevagye power is p= 350 W = AW - 100F ~ 250 wolts Pay At Qs