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m 36g 2. A mass of 3.00 kg is on an 35.0 incline. Tou It gul-3g in 35° = 16.8631 N a) If the mass is released from rest while on the incline, what will its speed be after traveling 1.50 m down the incline. Assume that the system is frictionless here. VO Vp - 1 a-S.62mk 39.-11.8631 Ar-1.50m 3 3 a = 5.62 m VW + 2 (5.62 mlc)(1.50m) 14. Ilms b) If the mass is released from rest while on the incline and its acceleration is 2.80 ms down the incline, what must the coefficient of kinetic friction be? Fon Fgs l = 3 g cos 35 = 24,0831 fx = x(24.0831), Eta, Fyn - fx = ma 161863)- MX (24.0831) - 3(2,87 mi) 16,8631-8,4 = Mx (24.0831) 8.4631 - Mk (24-01 31) MK 24.083) 24.0831 0.3514 = Mr
c) What magnitude applied force directed up the incline must act on the mass so that the acceleration produced is 2.80 m/s up the incline, if the system frictionless? IF-toppt tgv = 3(2.80) -Fapp cos 35° +16,8631N = 8,4 16.8691-874= Fapp cos 35° Lot 35 532 Lot 35° 10,335 Fapp x d) If a 40.0 N force directed up the incline acts on the mass and the resultant acceleration is 2.80m's up the incline, what must the coefficient of kinetic friction be? 1.8631 le ET Fgi -4000535 - M (24.0831)-3(260) 16.8631 - 40001 350-M024.0631) - 6:4 Στο $75 76-863)-4000135.-84 - M(24.0691) -24.3030 Mx 24.0631 -2 M = -1.009 =