Two blocks are positioned on surfaces, each inclined at an angle of 59.1 degrees with respect to the horizontal. The blocks are connected by a rope resting on a frictionless pulley at the top of the inclines, allowing them to slide together. The mass of the black block is 5.38 kg, and the coefficient of kinetic friction for both blocks and inclines is 0.360. Assume static friction has been overcome and that everything can slide.

What must be the mass of the white block if both blocks are to slide to the LEFT with an acceleration of [tex]1.5 \, \text{m/s}^2[/tex]?

A. 3.88 kg
B. 12.37 kg
C. 7.92 kg
D. 5.21 kg

This problem is a classic Physics example of Newton's laws and can be solved by setting up the appropriate equations of motion and friction. The black block on the left has a downward force due to gravity, given by F_black = m_black * g, where m_black is the mass of the black block and g is the acceleration due to gravity (9.8 m/s²). The force due to friction acting against this is f_black = m_black * g * cos(59.1) * µ_k, where µ_k is the coefficient of kinetic friction (0.360). Similarly, the white block on the right also has a force due to gravity and a force due to friction. If we want the two blocks to slide to the left, the net force acting to the left has to be equal to the total mass of the system times the acceleration (F_net = (m_black + m_white) * a), where a is the desired acceleration (1.5 m/s²). Solving these equations for m_white yields option b) 12.37 kg.

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