Question

Problem 2: Determine the velocity of the 60-lb block A if the two blocks are released from rest and the 40-lb block B moves 2 ft up the incline. The coefficient of kinetic friction between both blocks and the inclined planes is 0.10. 30r

Answer 1

Sol. Given that weight of Block (A) = Go lb weight of Block (B) = 40 lb Mk= 0 lo Stop- F.B.D of Block A conly) : --- Collo Goo 72T FA=liKNA Yaxis ↑ (+) NA TE daxis Apply The equation of motion Efy = mayo NA - 60 (Osgoo NA = 60 x 1/2 = 3016 NA = 3012 Then Friction force in Block(A) TFA = MK NA FA = 0.1030 FA = 316

Step- F.B.D of Block B only T holl Applying the equation of matiom 13.0 x a Efy=may=0 gi Ng -40 COD 300 =0 FB = UK NB NB = 2053 Jo FB2013 xoil = 253 F.B.D of full [ A Block 8 B Block] Step- T SB Golb KSA yolo THINA NA NB MK NB Applying the Principle of work and energy [Tz T J = {U1+2 0 B T2 = I MA VA + 1 MB UP VA₂= Velocity At Block A VA2 = Velocity At Black B whert

Ti= 2 mar + + mp up Where Vape in tial Velocity at Block A intial velocity at Block B assume that intial velocities at A an B Is Zoro VB1 So we So Vas VBI=0 Then 1 To EU ISA [Go Sin 600] -uosinoo °[ls] – 3521 23 x81 30 13°/S A-20581 - 3$ A+ 115 $8)) 2 EU2 Step-li find the value of SASSB 2SA + S8 = 1 [58= 2ft] giren 2 SA - - SB [ SA=IRt SB=2ft Because length Always positive we know that Velocity = Speed ult Time

2 SA= - SB Can be writtin in the form of VAS VB 2 VA = -VB 3 From Equation >> 30 33 x 1 x1 - 20x2-3x 1 213x2 = MAVA + I move =) 2.033 x60 (-2VA)? s VAZ ² - + Lx yo x 3२.५ 32.4 2.033 3.395 VA 0.7738 fts * sa IVA from Eq. VB = - 2 VA - 2x0.7738 Vg = -1.5476 475142