Question

2-A 30-kg block is connected by an inextensible cable through the pulley to the fixed surface, as shown in Figure. A 20 kg weight is attached to the pulley, which is free to move vertically. A force of magnitude given bellow tows the block. The system is released from rest at t= 0. P= 1000 N (a) What is the acceleration of the 30-kg block as a function of time? (b) How far does the block travel up the incline before it reaches a velocity of 60 cm/s? 30 kg 45° 1 = 0,3 20 kg

Answer 1

P=100t m, 30kgp u = 0.3 145 let m, -30 kg and my = 20 kg. let a, be the acceleration of block m, along the incline upward. 20 kg m2 let a, be the acceleration of block m2 (upward). i be the length of cable then 26, +1, + 3 = e. (value) Note that as blocks (m,, m2) move, 46 ls change but By remain same. Ž de dt = 0 as is constant value Since cable is inc I tensible + + dt dlz = =) + 2.dl, dt dt dt 4.de -vclocity of block m2 de acc. of block again differentiating 2.de + de dL² 22 =D dt² dt? block Similarly for m 2a + ag=0 acc is al laz/= 2loul acceleration of block m, is twice of block m, (az) (a, = 202) a = 2a. It t' be the tension in the cable.

Dynamic equilibaium of block m, des dived N acc.a, Р weight mig is divided/ along and perpendicular to the inclined plane mg sin45 т f mg mg cos 45 mg sin 45 em, gcos 45 mg f=MN 27y=0 0 (to to the plane) N = m, gcos 45 f = MN, M = coefficient of friction - f = M m, q cos 45 Na noimal reaction Along the plane EF=ma P-T-mgsin 45 - jum, gcos 45 = ma, I or block m₂ 2T - m2 g = m2 a EF-ma -> Az - T = M2/A2 + 9) 2 let a = a then a₂ = a - a = m, a a -m, g sin 45 -um , 9 205 45 = 2 -) P-M29m, g sin 45 - pimg cos 45º = a[m, + m2 2 P = 1000, 12-20, 9=9,81 m/s", m = 30 kg . H = 0-3 100+ - 30x9:!3044(***) - a (348) loot - 368.63198 = 35a

35 a = 1001 - 368.63198 a = I 35 100+ - 368.63198) a) .. acceleration of the 30 kg block along the inclinc (upward) is a = 100+ - 368.63193) 35 a = 2.957146-10-53234 a = 2.957/4t - 10.53234 m/s2 let depresent distance (travelled by block 30 kg) & V-ds represent velocity du = ds then a= dt dL dt du dt = 2.8571t - 10.5323 Sov = (2-85117 - 10:5323) dt V = 2.85710 -10.53 235 +4, 2 where is constant. AL time t=0, V = 0 0 = 0+, => 4,=0. velocity, v = 1.4285t-10.5323 = 1.4285 E - 10.5323 similarly V = ds dt 5 = 4285 -10.5323t +62 2 =0 as at t=0,s=0 S= 0.4761 +² -1 - 10-5323 672

1/= 60cm/s = v=0.6 m/s 0-6 = 1, 42957-10-53235 -> time, t = 7.4295 sec 2 -) 5 = 0.4761 % -10.53231 Substitute t = 7.4295 2 = = - 95.434 m. 30 kg block (moves down) by 95.4344 m when its velocity is 0.6m/s 31 (60cm) Initially, acca is -VC which means block moves downwards.) Es velocity v
I hope you understand. As the 30 kg block moves up then the 20 kg block also moves up. If the 30 kg block moves down then 20 kg also moves down. Initially system is at rest. So at time t=0, both velocity and displacement are taken as zero. Comment if you have any doubt.