Question

A 385-g tile hangs from one end of a string that goes over a pulley with a moment of inertia of 0.0125 kg ⋅ m2 and a radius of 15.0 cm. A mass of 710 g hangs from the other end of the string. When the tiles are released, the larger one accelerates downward while the lighter one accelerates upward. The pulley has no friction in its axle and turns without the string slipping. What is the tension in the string on the side of the 710-g tile?

Answer 1

the tension in the string is 5.59 N

Step-by-step explanation:

Here ,

m_1 = 0.385 Kg

m_2 = 0.710 Kg

Using second law of motion ,

a = F_net / effective mass

a = (0.710- 0.385)×9.8/(0.710 + 0.385 + 0.0125/0.15^2)

a = 1.93 m/s^2

Now , let tension be T ,

then,

mg-T=ma

0.710×g - T = 0.710×1.93

T = 5.59 N

the tension in the string is 5.59 N