Question

1) Mes my sits on top of the left upper edge of the mass m - 10.0 kg, which sits on a 100 m long ramp, 1) a. measured along the incline, that makes an angle of - 30.0 with the horizontal. A rope which wraps around a massless, frictionless pulley at the top of the ramp connects to my. The height of m is negligible (You can take its bight to essentially be o.), but it has a length of L -1.00 m m starts out with its center of a 7.00 m from the top of the ramp, measured along the incline, as shown in the figure You can take the size of as negligible. ic. You can take its height as zero, and its width as zero.) The coefficients of static and kinetic friction between , and the ramp are po-0.1550, and po-0.1450, respectively. The coeficients of static and kinetic friction between my and are - 0.1230, and -0.1000, respectively. The combinats of static and kinetic friction between my and the ramp are -0.1700, and = 0.1450, respectively Note that there are also bumpers at the top and bottom of the ramp that prevent the masses from going past those points. If the masses collide with the bumpers, they stop at that point. In other words, the s don't, for instance, bounce of the bumpers, They just come to an immediate stop.) The masses are initially released from rest. Measure all distances along the incline, from the top. pulley b umper 10.0 m Figure 1 - m is on top of the upper edge of the mass m - 100ks, which sits on a 100 m long rump, that makes an angle of B = 300 with the horizontalm is connected to my by a rope which wipe onda pulley at the top of the ramp. The length of mais 1.00 m and its center of mais initially 7.00 m from the top of the Tam. All other dimensions of the neglibble. The efficients of static and kinetic friction are th =0.1550, 0.1150. 2, 0.1250. = 1000 =0:1700 anda = 1150, where the script denote mi, denote maand denote the ramp (Not to scale) 10.0 kg a) What is the final resting place of the center of mass olm? b) What is the final resting place of c) How long after the nee d does it take me to get to its final resting place? d) How long after themes are released does it takes to get to its final resting place! e) What's the speed om just before arriving at its final resting place! 1) What's the speed of just before arriving at its final resting place? 50.0 kg m 3) What is the final resting place of the center of mass of b) What is the final resting place of 1) How long after them are released does it take to get to its finalisting place J) How long after the w a ter ed does it take me to get to its final resting place! k) What's the speed of just before arriving at its fissal resting place? 1) What's the speed of just before arriving at its final resting place?

please use significant figures for all the problems.

plz make sure all answer were in signficant and do all of them plz i havw this last photos

plz make sure all answer were in significant figuers and plz answer all of them

Answer 1

Soln for m₂ = 4.0.0 kg DI my megcaso 3 mag sina → frictim on a due tot i e fai mig sino mig caso Here it is given that m. = 10kg ie mass o = 3000 i.e angle 4105 & o. 1550 (static friction coeff 61wm, and ramp) Miok = 0.1450 (rinetic as 99 99 99 99 ] 4215 = 0.1250 (Static 9 9 9 m, and me). Miok = 0.1000 C Rinetec friction or coeff-blw m, and mo) 42os = 0.1700 (static 9 99 9 me andramp 4 201 = 0-1450 C kinetic 99 99 99 esse 3) taking m₂ = 40 kg we can see that m₂ >m, 1m₂ g sino > m, gsenol which means my will go downw and and my will come up ward. for m2 m2 g sino = 40 X 9 8 * sin 30° = 1196 N. Now we find all the friction forces

(1) friction of 2 due to I (mass ) = fals - 4215 m₂ g caso = (0.125) X 40 X 9.8 X iss 30° = 42.43 N. f2ik 4216 m₂ g Caso = Co.1000) X 640X9.8 los 3. 33.94 N (1) friction of m₂ on hamp. = 420k mag lasa = Co. 1450 ) x 40 x 9.8 x cas 30°. = 49.22 No (@feedban de kan lover o Now for mi + m, g sen o 9.8 X sin 3 0 = 10 X = 49 N for all friction forces. (1) friction of m, on the ramp fros = Mios (m. g case + m₂ g caso). - 0.155 X 50 X 98 XC = 65.77 N. flok = 110 ( m. g caso + Mzglas o) = 0.135 C mitm2) g caso = 0.145 x 50 x 9.8 x Ess 30° = 61.53 N. adro tobena friction when my is not on top. flok = 410 tim, tom) of caso in = 12.3058.N = 0.1458 9.88 iss 30X 16 -

have got all the values we needed. we have got a Now (a) Haefag efter Net downward force of mz>m, m₂ will ultimately hit downside bumper and m, will moves upward bcuz of it. length of m₂ = 1.00m M2 - 0-50m. to reach the downward bumper it has to cover distance = 10.00- 7-5o. = 3.50 m and m, cover distance = 3.50 m buz of it. So final resting resting place of com of m, E 3.50m) I (6) C final resting place of com of m₂ = 110.oo my cuz it touches bumper. time lico total time taken by mi = fal time taken to fall of of m₂ + taken on the ramp. å (i) time taken to fall of my. to a s= ut t I at?

C So S= 4 at ² → U12 > t = 25 (miton) / Fnet TE=ma l +(a e s=1 c relative displace on M₂ wrt m.) a= F net m, tmz a = (m₂ g sin o-f21k ) - (farkt mig sino + frok, 5o. (196-33.95) - (33.95 + 49 +12:31) a - = 1.33 mlsz. Since this is a real relative acc 18, total acc. . a = acc of m₂ + acc of m, q- 2x a = 2.66 misa. putting the values in above we get will be ao tra t= 12x1 2.66 = 0.867 8. Now time taken by me on ramp. s= ut + ļ at. Here uto becuz when my slips it has some velocity i.e. v² = u²+ zas v2 = zas

ince velocity and acc are to relative. V=2v and a = 2a. (20)2 = 2(2a) ci) 4v2 = 4a. 4 v2 = 4X 1.33 V = AX $T-33 V= 1.153 mis. So also s= ut t I at? 1.153 P a = Fret mitm2 a = (Magsino-fok)- m. g seno + frok). (m+m2) 4 when my is not 1960 - 49.223 - 49. - 12:30 top - 50 1.709 mis? a = S = 1.153+ + 1 (1.109) *+ 3 = 1.16 + + ļ (1-71) +². solving the 2 order ear we I t = 1.3148 get

ad de ading both the times will get. d) t = 0.867+ 1.314 t = 2:18 98V time taken by my will also be the same as time taken by my o'i t - 2:1818 time to arriving will be speed of m, just before V= ut at V= 1.16 + 1.171x t. V= 1.16+ 1-171 × 1.314. I= 3.407 mis.) (f) speed of me just before arriving will be same as m, 0 v= 3.407 mis /