Please, Read Question Below!

Question

Two objects are connected by a light string passing over a light, frictionless pulley as shown in the figure below. The object of mass m₁ = 6.60 kg is released from rest at a height h = 2.60 m above the table.

(a) Using the isolated system model, determine the speed of the object of mass m₂ = 3.00 kg just as the 6.60–kg object hits the table.

_____m/s

(b) Find the maximum height above the table to which the 3.00–kg object rises.

_____m



Answer 1

As the two masses are connected, the tensions in the string T and the acceleration a are the same.

The smaller mass will be accelerated up with a net force F_{net 2} = T - m2g. The larger mass will be accelerated down with a net force F_net1 = m1g - T. As T is the same for both masses, solve for T and equate the results.

But F_{net 2} = m2a and F_net1 = m1a

T = m2a + m2g = m1g - m1a

Substituting values and solving for a = 3.6 kg x 9.8m/s² / 9.6 kg = 3.68 m/s²

As the 3.00 kg weight will travel the same distance as m1 but

upward, the velocity can be found by

V = √(0 + 2 x a x 2.60 ) = 4.37 m/s

Please check math.