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The frictionless system shown is released from rest. After the right-hand mass has risen 75 cm, the object of mass 0.50m falls loose from the system. What is the speed of the right-hand mass when it returns to its original position?​

The frictionless system shown is released from rest. After the right-hand mass has-example-1

1 Answer

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Let
a be the acceleration of the masses. By Newton's second law, we have

• for the masses on the left,


1.3mg - T = 1.3ma

where
T is the magnitude of tension in the pulley cord, and

• for the mass on the right,


T - mg = ma

Eliminate
T to get


(1.3mg - T) + (T - mg) = 1.3ma + ma


0.3mg = 2.3ma


\implies a = (0.3)/(2.3)g \approx 0.13g \approx 1.3 (\rm m)/(\mathrm s^2)


Starting from rest and accelerating uniformly, the right-hand mass moves up 75 cm = 0.75 m and attains an upward velocity
v such that


v^2 = 2a(0.75\,\mathrm m) \\\\ \implies v \approx \sqrt{2\left(1.3(\rm m)/(\mathrm s^2)\right)(0.75\,\mathrm m)} \approx 1.4(\rm m)/(\rm s)

When the 0.5m mass is released, the new net force equations change to

• for the mass on the right,


mg - T' = ma'

where
T' and
a' are still tension and acceleration, but not having the same magnitude as before the mass was removed; and

• for the mass on the left,


T' - 0.8mg = 0.8ma'

Eliminate
T'.


(mg - T') + (T' - 0.8mg) = ma' + 0.8ma'


0.2mg = 1.8 ma'


\implies a' = (0.2)/(1.8)g = \frac19 g \approx 1.1(\rm m)/(\mathrm s^2)

Now, the right-hand mass has an initial upward velocity of
v, but we're now treating down as the positive direction. As it returns to its starting position, its speed
v' at that point is such that


{v'}^2 - v^2 = 2a'(0.75\,\mathrm m) \\\\ \implies v' \approx \sqrt{\left(1.4(\rm m)/(\rm s)\right)^2 + 2\left(1.1(\rm m)/(\mathrm s^2)\right)(0.75\,\mathrm m)} \approx \boxed{1.9(\rm m)/(\rm s)}

User Jafar Rasooli
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