Block A (Mass = 2.033 kg) and Block B (Mass = 1.680 kg) are attached by a massless string as shown in the diagram.

Question

Block A (Mass = 2.033 kg) and Block B (Mass = 1.680 kg) are attached by a massless string as shown in the diagram.

Block A sits on a horizontal tabletop. There is friction between the surface and Block A. The string passes over (you guessed it) a frictionless, massless pulley. Block B hangs down vertically as shown. When the two blocks are released, Block B accelerates downward at a rate of 3.310 m/s². 

What is the tension in the string?

What is the magnitude of the force of friction acting on Block A?

What is the coefficient of friction between the tabletop and Block A?

Answer 1

Mass B is accelerating downward so there is a net force equal to 1.680kg x 3.310 m/s². This net force is equal to the resultant of the downward force of gravity minus the tension in the string (T).

1.680kg x 3.310m/s² = 1.680kg x 9.807 m/s² - T

Solve for T to get the tension in the string. T = 10.90 N.

Mass B must have the same acceleration of mass A as they are attached. It's net force is 2.033 kg x 3.310 m/s². Therefore the net force is equal to T - f_r where f_r is the force of friction.

2.033 kg x 3.310 m/s² = 10.90 N - f_r. Solve for f_r = 10.90 N - 2.033kg x 3.310 m/s². = 4.171 N

As f_r = 4.171 N = μ x 2.033 kg x 9.807 m/s² , solving for μ = 0.2092.

Check all calculations.