Question 4 [25 points] Tension and friction Two boxes of the same mass m - 2 kg

Question

Question 4 [25 points] Tension and friction Two boxes of the same mass m - 2 kg are connected with a massless rope, and they are placed on a surface inclined at a 30 degrees above horizontal, as it is shown in the picture below Floor Between Block 1 and the inclined plane there is friction. The coefficiento of kinetic friction is = 0.5. Between block 2 and the surface there is no friction. a. Draw a free body diagram of the Block 1, of the rope, and of block 2, labeling carefully the forces. b. Compute the total force acting on Block 1, on block 2 and on the rope. Indicate magnitude and direction. c. Compute the acceleration of the system. d. If block 2 starts from rest at a distance d from the ground, what will be its final speed when it hits the bottom of the inclined plane?

Explanation / Answer

given,

mass = 2 kg

angle = 30 degree

coefficient of kinetic friction = 0.5

force in block 2

ma = mg * sin(30) - T

force in block 1

ma = T + mg * sin(30) - mu * mg * cos(30)

so,

subtracting both the equation

0 = T + mg * sin(30) - mu * mg * cos(30) - mg * sin(30) + T

0 = 2T - 0.5 * 2 * 9.8 cos(30)

T = 4.24352 N

Tension in the rope = 4.24352 N

force in block 2 = mg * sin(30) - T

force in block 2 = 2 * 9.8 * sin(30) - 4.24352

force in block 2 = 5.5565 N

force in block 1 = T + mg * sin(30) - mu * mg * cos(30)

force in block 1 = 4.24352 + 2 * 9.8 * sin(30) - 0.5 * 2 * 9.8 * cos(30)

force in block 1 = 5.5565 N

acceleration = force / mass

acceleration = 5.5565 / 2

acceleration = 2.77825 m/s^2

v^2 = 0 + 2 * 2.77825 * d

final speed v = sqrt(5.5565 * d)

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