A 5.7- kg concrete block rests on a level table. A 4.0- kg mass is attached to the block by a string passing over a light, frictionless pulley. If the acceleration of the block is measured to be 1.0 m/s2, what is the coefficient of friction between the block and the table?
See my answer to a later post of the same question.
To find the coefficient of friction between the block and the table, we need to use Newton's second law of motion.
First, let's analyze the forces acting on the system.
1. The weight of the concrete block (W1) is equal to its mass (m1) multiplied by the acceleration due to gravity (g):
W1 = m1 * g
2. The tension in the string (T) is equal to the weight of the attached mass (W2):
T = W2 = m2 * g
3. The force of friction (f) between the block and the table resists the motion and acts opposite to the applied force.
Since the system is in equilibrium, the force of friction (f) is equal to the tension in the string (T), but in the opposite direction:
f = T
Using Newton's second law, we can write the equation of motion for the block:
f = m1 * a
Substituting the expressions for f and T:
m2 * g = m1 * a
We can rearrange this equation to isolate the coefficient of friction (μ):
μ = (m2 * g) / (m1 * a)
Now, let's substitute the given values:
m1 = 5.7 kg (mass of the concrete block)
m2 = 4.0 kg (mass attached to the block)
a = 1.0 m/s^2 (acceleration of the block)
The acceleration due to gravity (g) is approximately 9.8 m/s^2.
μ = (4.0 kg * 9.8 m/s^2) / (5.7 kg * 1.0 m/s^2)
Calculating this expression:
μ ≈ 6.84 / 5.7
μ ≈ 1.199
Therefore, the coefficient of friction between the block and the table is approximately 1.199.