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In the Atwood machine shown below, m1 = 2.00 kg and m2 = 7.60 kg. The masses of

ID: 1528409 • Letter: I

Question

In the Atwood machine shown below, m1 = 2.00 kg and m2 = 7.60 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 = 2.30 m/s downward.

(a) What is the magnitude of the acceleration of m1?
1 m/s2
(b) How far will m1 descend below its initial level?
2 m

(c) Find the velocity of m1 after 1.80 s.

magnitude     3 m/s direction 4---Select---downwardupwardvelocity is zero m m9 2

Explanation / Answer


atwood's machine


for block1

T - m1g = m1a

T = m1g + m1a...........(1)


for block 2

m2g - T = m2a..........(2)

substituting 1 in 2


m2g - m1g - m1a = m2a


m2g - m1g = m1a + m2a

g*(m2-m1) = a*(m1+m2)

acceleration , a = (m2-m1)*g/(m1+m2)

(a)


acceleration a = (7.6-2)*9.8/(7.6+2) = 5.72 m/s^2

(b)

vf^2 - vi^2 = 2*a*y

0 - 2.3^2 = -2*5.72*y

y = 0.46 m

(c)

v = vi + a*t


v = -2.3 + (5.2*1.8)


magnitude = 7.06 m/s

direction = upwards