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

Answer 1

acceleration of lighter block= a= (2-6)g/(6+2) = -4.9 m/s^2

initial speed = u = 2.2 m/s,

a.

Now,
v^2 = u^2 - 2as
0 = 2.2^2 - 2 x 4.9 x s

Distance = s= 0.49388 meters

b.
time for velocity to be zero,
0 = 2.2 - 4.9 t
time = t = 0.449 sec

Now, m1 will move upwards,
u = 0
acceleration = 4.9 m/s^2
time = t' = 1.8 - 0.449 = 1.351 sec
speed = v = u+at
= 0 + 1.351 x 4.9
= 6.62 m/s