A string going over a massless frictionless pulley connects two blocks of masses 6.4 kg and 13 kg. As shown on the picture below, the 6.4 kg block lies on a 32◦ incline; the coefficient of kinetic friction between the block and the incline is μ= 0.3. The 13 kg block is hanging in the air. The 13 kg block accelerates downward
while the 6.4 kg block goes up the incline
with the same acceleration.
Given
g= 9.8 m/s^2, what is the acceleration of the system?
Answer in units of m/s^2.
To find the acceleration of the system, we need to analyze the forces acting on each block separately.
Let's start with the 13 kg block hanging in the air. The only force acting on this block is its weight, which is given by:
Weight = mass * gravitational acceleration
Weight = 13 kg * 9.8 m/s^2 = 127.4 N
Now let's consider the 6.4 kg block on the incline. The forces acting on this block are its weight, the normal force, and the frictional force.
The weight component parallel to the incline can be calculated as follows:
Weight_parallel = weight * sin(angle of incline)
Weight_parallel = 6.4 kg * 9.8 m/s^2 * sin(32°) ≈ 33.05 N
The normal force can be calculated as follows:
Normal Force = weight * cos(angle of incline)
Normal Force = 6.4 kg * 9.8 m/s^2 * cos(32°) ≈ 54.85 N
The frictional force can be calculated as follows:
Frictional Force = coefficient of kinetic friction * normal force
Frictional Force = 0.3 * 54.85 N ≈ 16.45 N
Since the 13 kg block is accelerating downward and the 6.4 kg block is accelerating upward with the same acceleration, we can set up the following equation of motion for the system:
Net force = (13 kg + 6.4 kg) * acceleration
The net force is given by:
Net force = Weight_13kg - Weight_parallel_6.4kg + Frictional Force_6.4kg
Net force = 127.4 N - 33.05 N + 16.45 N
Net force = 110.8 N
Now we can solve for the acceleration:
110.8 N = (13 kg + 6.4 kg) * acceleration
110.8 N = 19.4 kg * acceleration
Simplifying the equation:
acceleration = 110.8 N / 19.4 kg ≈ 5.72 m/s^2
Therefore, the acceleration of the system is approximately 5.72 m/s^2.
To find the acceleration of the system, we need to analyze the forces acting on each block and apply Newton's second law of motion.
Let's start with the 13 kg block. The only force acting on it is its weight, which is given by the formula:
Weight = mass * acceleration due to gravity
Weight = 13 kg * 9.8 m/s^2 = 127.4 N (Newtons)
Since the 13 kg block is accelerating downward, we can write the equation:
Force_down - Force_friction = mass * acceleration
The force down is equal to the weight of the block (127.4 N). The force of friction can be calculated by multiplying the coefficient of kinetic friction (μ) by the normal force, where the normal force is the force perpendicular to the incline.
Force_friction = μ * Normal force
The normal force is equal to the weight of the block times the cosine of the angle of the incline (32 degrees).
Normal force = weight * cos(angle) = 127.4 N * cos(32 degrees) = 107.9 N
Substituting the values into the equation, we have:
127.4 N - μ * Normal force = 13 kg * acceleration
127.4 N - (0.3) * 107.9 N = 13 kg * acceleration
Now, let's move on to the 6.4 kg block on the inclined plane. The forces acting on this block are its weight, the force of friction, and the force pulling it up the incline.
The weight of the block is given by:
Weight = mass * acceleration due to gravity
Weight = 6.4 kg * 9.8 m/s^2 = 62.72 N
The force of friction is given by:
Force_friction = μ * Normal force
Again, the normal force is equal to the weight of the block times the cosine of the angle of the incline.
Normal force = weight * cos(angle) = 62.72 N * cos(32 degrees) = 53.17 N
The force pulling the 6.4 kg block up the incline is equal to the mass of the block times the acceleration of the system.
Force_up = 6.4 kg * acceleration
Now, let's write the equation for the 6.4 kg block:
Force_up - Force_friction - Force_down = mass * acceleration
Substituting the values, we have:
6.4 kg * acceleration - μ * Normal force - 62.72 N = 6.4 kg * acceleration
Now we have two equations with two unknowns (Force_down and acceleration). We can solve them simultaneously.
From the equation for the 13 kg block:
127.4 N - (0.3) * 107.9 N = 13 kg * acceleration
From the equation for the 6.4 kg block:
6.4 kg * acceleration - (0.3) * 53.17 N - 62.72 N = 6.4 kg * acceleration
Simplifying the equations, we have:
34.53 N = 13 kg * acceleration
4.54 N - 20.2 N = 6.4 kg * acceleration
Solving these equations, we find the acceleration of the system:
acceleration = (34.53 N / 13 kg) = 2.66 m/s^2