A 12 kg mass on a horizontal friction-free air track is accelerated by a string attached to a 100 kg mass hanging vertically from a pulley as shown. Compare the accelerations when the masses are interchanged, that is, for the case where the 100 kg mass dangles over the pulley, and then for the case where the 12 kg mass dangles over the pulley.
acceleration with 100 kg mass dangling g
acceleration with 12 kg mass dangling g
What does this indicate about the maximum acceleration of such a system of masses?
The pulling mass is the hanging weight, the accelerating mass is the sum of the pulling mass + track mass.
F=ma
Pullingweight=acceleratingmass*acceleration.
Remember that the pulling weight is the hanging mass*g.
To compare the accelerations when the masses are interchanged, we can start by setting up the equations using Newton's second law (F = ma):
When the 100 kg mass dangles over the pulley:
The force on the 100 kg mass is given by F = (mass of the 100 kg mass) * g, where g is the acceleration due to gravity (9.8 m/s^2). This force is equal to the mass of the system (100 kg + 12 kg) multiplied by the acceleration.
So we have: (100 kg) * g = (100 kg + 12 kg) * acceleration
When the 12 kg mass dangles over the pulley:
The force on the 12 kg mass is given by F = (mass of the 12 kg mass) * g. This force is equal to the mass of the system (12 kg + 100 kg) multiplied by the acceleration.
So we have: (12 kg) * g = (12 kg + 100 kg) * acceleration
Now, let's solve for acceleration in both cases:
For the case with the 100 kg mass: acceleration = (100 kg) * g / (100 kg + 12 kg)
For the case with the 12 kg mass: acceleration = (12 kg) * g / (12 kg + 100 kg)
Now we can calculate the values:
Using g = 9.8 m/s^2, we have:
For the case with the 100 kg mass: acceleration = (100 kg) * 9.8 m/s^2 / (100 kg + 12 kg) ≈ 8.36 m/s^2
For the case with the 12 kg mass: acceleration = (12 kg) * 9.8 m/s^2 / (12 kg + 100 kg) ≈ 0.94 m/s^2
So, the acceleration with the 100 kg mass dangling is approximately 8.36 m/s^2, while the acceleration with the 12 kg mass dangling is approximately 0.94 m/s^2.
This indicates that the maximum acceleration of the system depends on the mass of the pulling weight. In this case, with a 100 kg pulling weight, the maximum acceleration is higher compared to a 12 kg pulling weight. So, the greater the mass of the pulling weight, the greater the maximum acceleration of the system.
In the given scenario, we have a 12 kg mass on a horizontal friction-free air track, and a 100 kg mass hanging vertically from a pulley. We need to compare the accelerations when the masses are interchanged, meaning that we will compare the acceleration when the 100 kg mass dangles over the pulley and when the 12 kg mass dangles over the pulley.
To analyze this, let's first consider the acceleration when the 100 kg mass is dangling over the pulley. According to Newton's second law (F = ma), the pulling weight (the force exerted by the hanging mass) is equal to the accelerating mass times acceleration. The pulling weight is given by the hanging mass (100 kg) multiplied by the acceleration due to gravity (g). Therefore, the equation can be written as:
Pulling weight = accelerating mass * acceleration
(100 kg) * (g) = (accelerating mass) * (acceleration)
Now, let's consider the acceleration when the 12 kg mass is dangling over the pulley. Again, using the same equation, we have:
Pulling weight = accelerating mass * acceleration
(12 kg) * (g) = (accelerating mass) * (acceleration)
Comparing the two equations, we can see that the only difference is the value of the accelerating mass. In both cases, the pulling weight is equal to the gravitational force (mass * g). Therefore, we can conclude that the acceleration is inversely proportional to the mass.
This indicates that the maximum acceleration of the system of masses will be larger when the lighter mass (12 kg) is dangling over the pulley, compared to when the heavier mass (100 kg) is dangling over the pulley. In other words, the lighter mass will experience a greater acceleration compared to the heavier mass.