Two blocks are connected over a massless, frictionless pulley. The mass of block 2 is 8.00 kg, and the coefficient of kinetic friction between block 2 and the incline is 0.250. The angle θ of the incline is 28.0°. Block 2 slides down the incline at constant speed. What is the mass of block 1? (Express your answer to three significant figures.)

Question

Two blocks are connected over a massless, frictionless pulley. The mass of block 2 is 8.00 kg, and the coefficient of kinetic friction between block 2 and the incline is 0.250. The angle θ of the incline is 28.0°. Block 2 slides down the incline at constant speed. What is the mass of block 1? (Express your answer to three significant figures.)

Answer 1

To find the mass of block 1, we can use the concept of forces and motion.

First, let's analyze the forces acting on the system.

For block 2, the forces acting on it are:
1. The force of gravity (mg), which acts vertically downwards.
2. The normal force (N2), which acts perpendicular to the incline.
3. The force of friction (f2), which acts opposite to the direction of motion.

For block 1, since it is connected to block 2 and there is no slipping or tension, the force acting on it is only the force of gravity (m1g), which acts vertically downwards.

Since block 2 is sliding down the incline at a constant speed, the net force acting on it in the direction of motion is zero.

Using Newton's second law (F = ma), the forces in the direction of motion can be written as:

For block 2:
mg sin(θ) - f2 = 0, where sin(θ) is the component of the force of gravity acting down the incline.

For block 1:
m1g = 0, since there is no acceleration in the vertical direction.

We can rearrange the equation for block 2 to solve for f2:
f2 = mg sin(θ)

The force of friction, f2, can be calculated using the coefficient of kinetic friction (μ) and the normal force (N2) as follows:
f2 = μ N2

The normal force (N2) can be determined by balancing the forces perpendicular to the incline:
N2 = mg cos(θ)

Substituting this value of N2 into the equation for f2, we get:
f2 = μ mg cos(θ)

Now we can substitute the value of f2 into the equation for block 2:
mg sin(θ) - μ mg cos(θ) = 0

Simplifying, we get:
sin(θ) - μ cos(θ) = 0

Now we can solve for the mass of block 1 by dividing both sides of the equation by m1g:
sin(θ)/m1g - μ cos(θ)/m1g = 0

Finally, isolating m1, we get:
m1 = sin(θ) / (μ cos(θ))

Now we can substitute the given values into the equation:
θ = 28.0°
μ = 0.250

Calculating m1, we get:
m1 = sin(28.0°) / (0.250 * cos(28.0°))

m1 ≈ 6.512 kg (rounded to three significant figures)

Therefore, the mass of block 1 is approximately 6.512 kg.