Two blocks,m₁ = 1.0 kg and m₂ = 2.0 kg,are connected by a light string as shown in the figure.If the radius of the pulley is 1.0 m and its moment of inertia is 5.0 kg.m²,the acceleration of the system is A)(1/6)g B)(3/8)g C)(1/8)g D)(1/2)g E)(5/8)g

Question

Quizplus · Accepted Answer

Since the two blocks are connected by a light string, the tension in the string is the same on both sides of the pulley. Let T be the tension in the string.
For m1 = 1.0 kg, the force equation is: 
T - m1g = m1a

For m2 = 2.0 kg, the force equation is:
m2g - T = m2a

Solving the above two equations simultaneously, we get:
a =(m2-m1)g/(m1+m2)=g/8=0.125g

Therefore, the acceleration of the system is (1/8)g, which corresponds to option C.

Two Blocks,m1 = 1

Two Blocks,m₁ = 1