Block A (Mass = 2.319 kg) and Block B (Mass = 1.870 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 2.250 m/s2.

The Tension in the string is 14.1 N.

b.)What is the magnitude of the force of friction acting on Block A?
c.) What is the coefficient of friction between the tabletop and Block A?

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To find the magnitude of the force of friction acting on Block A, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the net force acting on Block A is the force of friction, and the acceleration is the acceleration of Block B.

b) To calculate the force of friction, we can use the equation:

Force of friction (f) = mass of Block A (m) * acceleration of Block B (a)

Let's plug in the values given in the problem:

Mass of Block A (m) = 2.319 kg
Acceleration of Block B (a) = 2.250 m/s^2

f = 2.319 kg * 2.250 m/s^2
f ≈ 5.218 N

Therefore, the magnitude of the force of friction acting on Block A is approximately 5.218 N.

c) To find the coefficient of friction between the tabletop and Block A, we can use the equation:

Coefficient of friction (μ) = force of friction (f) / normal force (N)

The normal force is the force exerted by the tabletop on Block A perpendicular to the surface.

Now, we know that the tension in the string is 14.1 N, and this tension is equal to the force of gravity acting on Block B, which is also the normal force acting on Block A. So we can say:

Normal force (N) = tension in the string = 14.1 N

Plugging in the values:

Coefficient of friction (μ) = 5.218 N / 14.1 N
μ ≈ 0.370

Therefore, the coefficient of friction between the tabletop and Block A is approximately 0.370.