Question

7. [-/2 Points] DETAILS SERPSE 10 10.5.OP.017. MY NOTES PRACTICE ANOTHER Two boxes are connected to a cord, and the cord is hung over a pulley connected to the ceiling, as shown in the figure below. MOR m mo The masses of the boxes are m = 16.0 kg and m2 = 11.0 kg, the mass of the pulley is M = 5.00 kg, and the radius of the pulley is R = 0.300 m. Box mz is initially on the floor, and box m, is initially 4.70 m above the floor when it is released from rest. The pulley's axis has negligible friction. The mass of the cord is small enough to be ignored, and the cord does not slip on the pulley, nor does it stretch. (a) How much time (in s) does it take box m, to hit the floor after being released? At = S (b) How would your answer to part (a) change if the mass of the pulley were neglected? (Enter the time, in seconds, it takes box m, to hit the floor if the mass of the pulley were neglected.) Atz = S Need Help? Read It

Answer 1

Δt¹= 2.38s.

Δt²= 2.28s.

lule Massim We have given mi= 16 kg m2 = 11 kg Mass of pulley= 5kg Tz A Tz Radius of pulley = 0.300m TITT2 Wir mig W2=M2g for mass mi T2 - Mag=maa for mass my mig-Ti = mia UD) for fulley of Mars (m) : Torque = IX e= IX = (T.-T2R and a = Rx and I= 12 mR² (Ti-Ta) R = 4met (3) (T.T2) R = 1 mg To-T2= ma 2 on Adding er sis and equii) T2-mag + mig-Ti = m,atma (T2-T) +(mi-m)4- (mitm) a - ma + (mi-m) 9- (mitm) a (mı-m) f = (mitm)a+Ma af mitme *] a- (mi-mi) g (16-11) X98 [ Clube +M/2) (16+11+ 5/2) m/s 2 2 2 ) ~ 1.66 (mith2 +

① Now we have calculated the acceleration of System az 166m/52 for massmi: distance abone from floor = 4.7om time we have to calculated = ? &= ut Initially u=0 ut that? 470= 0+1 at² ath t = 24 4.70= at² 2 t= 2x4.70 n 2.38 sec - 1.66 Ati ~ 2.38 sec ** If mass of pally were neglected Ti = T2 for massme: Now time should be T-M2g = mga for massim, mig-T= ma t= on Adding - 2x4.70 1.815 ř 2.276 sec Ti-Mag + mig-T = maatmia & t-magtmig-1 = a(mitm) (mi-m) g = (mitm) a Ataň 2.28 sec 2) [ (m, -m2) g) mitm (16+11) 1.815 m/s It

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