In an Atwood\'s machine , one block has a mass of 600.0 g , and the other a mass

Question

In an Atwood's machine, one block has a mass of 600.0 g, and the other a mass of 715.0 g. The pulley, which is mounted in horizontal frictionless bearings, has a radius of 5.90 cm. When released from rest, the heavier block is observed to fall 61.8 cmin 2.34 s (without the string slipping on the pulley).

What is the magnitude of the acceleration of the 600.0-g block?

What is the magnitude of the acceleration of the 715.0-g block?

What is the magnitude of the tension in the part of the cord that supports the 600.0-g block?

What is the magnitude of the tension in the part of the cord that supports the 715.0-g block?

What is the magnitude of the angular acceleration of the pulley?

What is the rotational inertia of the pulley?

What is the change in the potential energy of the system after 2.34 s?

Explanation / Answer

Part A)

a = 2h/t2

h = 61.8 cm

t = 2.34 s

magnitude of the acceleration of the 600.0-g block, a = 0.226 m/s2

Part B)

In atwood's machine, both the masses are attached to the common string, hence the magnitudes of acceleration of both the masses would be same.

Hence, magnitude of the acceleration of the 715.0-g block, a = 0.226 m/s2

Part C)

the magnitude of the tension in the part of the cord that supports the 600.0-g block is given by

T1 = m(g + a)

= 0.6*(9.81 + 0.226)

T1 = 6.0216 N

Part D)

the magnitude of the tension in the part of the cord that supports the 715.0-g block is given by

T2 = M(g - a)

= 0.715*(9.81 - 0.226)

T2 = 6.85 N

Part E)

the magnitude of the angular acceleration of the pulley is given by

? = a/r

= 0.226/(5.9*10-2)

? = 3.83 rad/s2

Part F)

the rotational inertia of the pulley is given by

I = (T2 - T1)r/?

= (6.85 - 6.0216)*(5.9*10-2)/3.83

I = 0.0128 kgm2

Part G)

the change in the potential energy of the system after 2.34 s is given by, mgh - Mgh

(m - M)gh = (0.6 - 0.715)*9.81*61.8*10-2

Change in the potential energy of the system = -0.697 J

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