In the Atwood machine shown below, m1 = 2.00 kg and m2 = 6.00 kg. The masses of
ID: 1965595 • Letter: I
Question
In the Atwood machine shown below, m1 = 2.00 kg and m2 = 6.00 kg. The masses of the pulley and string are negligible by comparison. The pulley turns without friction and the string does not stretch. The lighter object is released with a sharp push that sets it into motion at vi = 2.20 m/s downward.(a) How far will m1 descend below its initial level?
1 m
In the Atwood machine shown below, m1 = 2.00 kg and m2 = 6.00 kg. The masses of the pulley and string are negligible by comparison. The pulley turns without friction and the string does not stretch. The lighter object is released with a sharp push that sets it into motion at vi = 2.20 m/s downward.
(a) How far will m1 descend below its initial level?
1 m
(b) Find the velocity of m1 after 1.80 s.
2 m/s 3 ---Direction--- upward downward velocity is zero
Explanation / Answer
1) They give you the initial velocity. The final velocity where the thing turns around is zero. Use your handy equation of motion: vi^2 = 2 a x Solve for the distance it will descend before it goes back: So x = vi^2 / 2a Plug in the formula for acceleration: x = vi^2 (m2+m1) / (2g (m2-m1) ) 2) final velocity = initial velocity + at They give you the formula for acceleration. Note that the acceleration is opposite the direction of the initial motion because you started the lighter one going down. I'll call the initial direction positive and the acceleration negative. vf = vi - g (m2-m1)/(m2+m1) t