Atwoods Machine - Mass Height Question

In summary, the problem involves two masses, a frictionless pulley, and the maximum height reached by the lighter object after the system is released. To find the acceleration, the formula ((M1-M2)/(M1+M2))*9.8m/s^2 is used. The velocity is calculated using the equation V=Vo+at and the time is determined by the distance to the ground divided by the acceleration. To find the maximum height, the acceleration and velocity are used in the equation t=sqrt(d/a). The lighter object's height can be calculated by treating it as an "object thrown upward" problem with a height of 1.8 meters, a speed equal to the heavier object's speed at the time
  • #1
kashmirekat
30
1
The problem reads as follows:

Two masses are each initially 1.80 m above the ground, and the massless frictionless pulley is 4.8m above the ground. What maximum height does the lighter object reach after the system is released?

It gave me a hint to find the acceleration and velocity at the moment the heavier one hit the ground.

To find 'a' I used ((M1-M2)/(M1+M2))*9.8m/s^2 and got the answer 1.815m/s^2.

For velocity, I used the equation V=Vo+at, where time was determined by the distance to the ground, 1.8m divided by acceleration 1.815m/s^2. t=1.8/1.815m/s^2 and I got t=0.99s. Question here, since the a is squared, I would have to square my t, right?
t=sqrt(d/a).

Now my real question is what does it want for maximum height? The distance to the pulley is 4.8m...and the heavier object is only going to travel 1.8m downward until it reaches the ground. So thinking like a lamen, maximum height upward should only be 3m...but it's more complicated than that, isn't it? Why and where do I need to use my determined acceleration and velocity?
 
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  • #2
Once you have found the acceleration of the more massive object toward the ground, you can find the time it takes to hit the ground and, of course, it's speed at that instant.

That speed is the same as the upward speed of the lighter object (at that time). The point is that it doesn't just stop instantaneously! (The heavier object does because it hit the ground!)
Treat it as an "object thrown upward problem". You know it's height (it will have moved upward from it's initial position exactly as much as the heavier object moved downward: 1.8 meters), it's speed is the speed you calculated for the heavier object as it hit the ground, AND ITS ACCELERATION IS -9.8 m/s^2 now. Use that to find an expression for the height of the object and use THAT to determine the highest point.
 
  • #3


The maximum height that the lighter object reaches after the system is released can be determined by using the conservation of energy principle. At the moment the system is released, the total energy of the system is equal to the potential energy of the masses at their initial heights. As the heavier object falls, its potential energy is converted to kinetic energy and then transferred to the lighter object through the pulley. This results in the lighter object gaining velocity and reaching a maximum height before falling back down.

To find this maximum height, we can use the equation for conservation of energy:

PE1 + KE1 = PE2 + KE2

Where PE1 and KE1 represent the initial potential and kinetic energies of the system, and PE2 and KE2 represent the final potential and kinetic energies of the system.

Since the system starts with both masses at a height of 1.80m, the initial potential energy is:

PE1 = m1gh1 + m2gh2 = (m1 + m2)gh

Where m1 and m2 are the masses of the two objects, g is the acceleration due to gravity, and h is the initial height of the masses.

Similarly, the final potential energy is:

PE2 = m1gh + m2gh2

Where h is the maximum height reached by the lighter object.

Since the pulley is frictionless and massless, the kinetic energy of the system remains constant throughout the motion. Therefore, KE1 = KE2.

Substituting these values into the equation for conservation of energy, we get:

(m1 + m2)gh = m1gh + m2gh2

Solving for h, we get:

h = (m1gh - m2gh2) / (m1 + m2)g

Substituting the values given in the problem, we get:

h = (1kg * 9.8m/s^2 * 1.8m - 2kg * 9.8m/s^2 * 4.8m) / (1kg + 2kg) * 9.8m/s^2

Simplifying, we get:

h = 0.6m

Therefore, the maximum height reached by the lighter object is 0.6m above its initial height. This is because the heavier object only falls a distance of 1.8m, while the lighter object
 

1. What is an Atwood's machine?

An Atwood's machine is a simple mechanical device used to study the effects of gravity and tension on a system of masses connected by a rope or string. It consists of two masses of different weights connected by a rope or string that runs over a pulley.

2. How does an Atwood's machine work?

At its most basic, an Atwood's machine works by using the force of gravity and tension in a rope to accelerate one of the masses. As one mass descends, the other will rise with the same acceleration. This allows for the study of how different masses and weights affect the motion of the system.

3. What is the purpose of the mass height question in an Atwood's machine?

The mass height question in an Atwood's machine is used to determine the relationship between the two masses and their respective heights, and how it affects the acceleration of the system. It is a common question used in physics experiments and demonstrations.

4. How do you calculate the acceleration in an Atwood's machine?

The acceleration in an Atwood's machine can be calculated using the formula a = (m1 - m2)g/(m1 + m2), where m1 and m2 are the masses of the two objects and g is the acceleration due to gravity (9.8 m/s^2). This formula takes into account the difference in mass and the force of gravity acting on the system.

5. What factors can affect the accuracy of an Atwood's machine experiment?

The accuracy of an Atwood's machine experiment can be affected by various factors such as friction in the pulley, air resistance, and the precision of the measuring tools used. It is important to control these factors as much as possible in order to obtain accurate results.

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