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?
To find the magnitude of the force of friction acting on Block A, we need to use Newton's second law of motion. The equation that relates force, mass, and acceleration is F = ma.
First, we need to calculate the force acting on Block A. This force is equal to the tension in the string. We are given that the tension is 14.1 N.
F = 14.1 N
Next, we need to calculate the mass of Block A, which is given as 2.319 kg.
m = 2.319 kg
Using Newton's second law, we can rearrange the formula to find the acceleration:
F = ma
Solve for acceleration:
a = F / m
= 14.1 N / 2.319 kg
≈ 6.08 m/s^2
The acceleration of Block A is approximately 6.08 m/s^2.
Now, we can find the force of friction acting on Block A. The force of friction can be calculated using the formula:
f_friction = μ * N
Where μ is the coefficient of friction and N is the normal force. In this case, the normal force is equal to the weight of Block A, which is given by:
N = m * g
Where g is the acceleration due to gravity, which is approximately 9.8 m/s^2.
Substituting the values, we get:
N = 2.319 kg * 9.8 m/s^2
≈ 22.72 N
Therefore, the normal force N is approximately 22.72 N.
Finally, we can calculate the force of friction:
f_friction = μ * N
To find the coefficient of friction, we rearrange the formula:
μ = f_friction / N
Substituting the values, we have:
μ = F_friction / N
= m * a / N
= (2.319 kg * 6.08 m/s^2) / 22.72 N
≈ 0.620
Therefore, the coefficient of friction between the tabletop and Block A is approximately 0.620.