Two blocks are positioned on surfaces, each inclined at the same angle of 59.1 degrees with respect to the horizontal. The blocks are connected by a rope which rests on a frictionless pulley at the top of the inclines as shown, so the blocks can 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 is must be the mass of the white block if both blocks are to slide to the LEFT at an acceleration of 1.5 m/s^2?

3.88 kg

12.37 kg

7.92 kg

5.21 kg

Through application of Newton's Second Law and force calculations, it is found that the mass of the white block must be 7.92 kg to enable both blocks to slide to the left.

Explanation:

To solve this problem, we need to apply Newton's Second Law in the form of ΣF_net = ma. ΣF_net is the sum of all forces along the direction of motion, m is the mass, and a is the acceleration of the system. For the black block, the net force in the direction of motion (which is downhill) is the sum of the force of gravity and the kinetic frictional force against it. The force of gravity can be broken down into its components, where the gravity force down the plane is mgsin(θ) and the force normal to the plane is mgcos(θ). The kinetic friction force can be calculated as μkFN where FN is the normal force (mgcos(θ)). Now, for the white block, it must be heavy enough to pull the black block upwards against its friction and its component of weight down the slope.

By setting up the equations and calculating, the mass of the white block can be found to be 7.92 kg which is option 3.

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