Question

The Atwood's machine has two hanging masses. The mass of the A is  3m and the mass B is 2m . The pulley C can be considered as a solid cylinder with mass  m and radius R . There is no slipping between the cord and the surface of the wheel. Here,  m= 2.00 kg,  2m= 4.00 kg, and  3m= 6.00 kg.

(Figure 1)

Part A

Part complete

What is the speed of mass A after it falls 2.10 m down?

v(f)=…..m/s

Answer 1

solution: > —$ mp3 < I figure o 1) АЯ figure ③ figure ③. * For . figure figure o o ТАЯ - TA - да ТА - mA ( 3-o) -> 6) of for figure ③ . Те - mp3 - mea Тв- по(+а) ->ы. * For figure ③, pulley В (ТА-Тв) = E-mp Re) 1 TA-TO= 4 Mpa de from equation @ and 6 , we get (ть 3 - +a) - Фея + ®ga ) = Мра h 4 – mg 3 = (A++ х) 4 (sm) 3-(27)3 = (эт) +2m+ 2)

mg=(5.5) ma a = 9 9.81 >> 178 milya Sur of Using the equation V2_No2zzas * here No 20 mls so 2.10 m & Then Ne Vaas >> V281.98 x2:10 :.1V4 = 2.434 mls|