In the Atwood's machine of Figure 8-23, the two masses shown are initially at rest at the same height. After they are released, the large mass, m2, falls through a height h and hits the floor, and the small mass, m1, rises through a height h. The mass m2 remains at rest once it hits the floor, but the mass m1 continues moving upward. How much higher does m1 go after m2 has landed? Give your answer for the case h = 1.4 m, m1 = 3.4 kg, and m2 = 4.1 kg.
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Well, let me tell you, this Atwood's machine is quite a show! It's like a game of see-saw gone wrong. Now, when m2 falls down, it's like saying "See ya later, Alligator!" and hits the floor. But poor m1 still wants to reach for the sky, not giving up so easily.
Now, to figure out how high m1 goes, we can use some good old conservation of energy. Remember, energy is a little like money, it's gotta be conserved, otherwise, we'll have budget cuts!
Here's the deal: The potential energy gained by the falling mass, m2, is equal to the potential energy gained by the rising mass, m1. It's like a fair trade, you know?
So, we have m2g × h = m1g × Δh, where g is the acceleration due to gravity. Now we plug in the numbers: h = 1.4 m, m1 = 3.4 kg, and m2 = 4.1 kg. Crunching the numbers, we find:
4.1 kg × 9.8 m/s² × 1.4 m = 3.4 kg × 9.8 m/s² × Δh
Solving the equation, we find Δh is approximately 2.092 m. So, m1 travels an additional 2.092 meters higher after m2 has landed.
Isn't physics just marvelous? It's like a circus act, but with numbers instead of acrobats!
To determine how much higher m1 goes after m2 has landed, we can use the conservation of mechanical energy.
The initial mechanical energy of the system is equal to the final mechanical energy of the system.
Initial mechanical energy = Potential energy of m1 + Potential energy of m2
Final mechanical energy = Potential energy of m1' (after m2 has landed) + Potential energy of m2' (at rest on the floor)
Since both masses start at rest at the same height, their initial potential energies are equal.
Initial mechanical energy = Potential energy of m1 + Potential energy of m2
= m1 * g * h + m2 * g * h
= (3.4 kg) * (9.8 m/s^2) * (1.4 m) + (4.1 kg) * (9.8 m/s^2) * (1.4 m)
= 46.694 J + 57.2444 J
= 103.9384 J
After m2 has landed, all of its potential energy is converted into kinetic energy.
Final mechanical energy = Potential energy of m1' + Potential energy of m2'
= 0 + m1' * g * h'
To find h' (how much higher m1 goes after m2 has landed), we can set up the equation:
Initial mechanical energy = Final mechanical energy
103.9384 J = 0 + (3.4 kg) * (9.8 m/s^2) * h'
We can now solve for h':
103.9384 J = 33.32 kg * m^2 / s^2 * h'
h' = 103.9384 J / (33.32 kg * m^2 / s^2)
h' ≈ 3.125 m
Therefore, m1 goes approximately 3.125 meters higher after m2 has landed.
To begin, let's identify the key principles involved in this problem. The Atwood's machine operates under the principles of conservation of energy and Newton's laws of motion.
First, let's consider the motion of the system after the masses are released. As mass m2 falls through height h and hits the floor, it comes to a stop. At this point, all of the potential energy of m2 has been converted into other forms of energy (such as kinetic energy and heat).
Now, let's focus on mass m1. As m2 falls, it transfers some of its potential energy to m1 through the tension force in the string connecting the masses. This causes m1 to rise. The work done by the tension force on m1 results in an increase in m1's kinetic and potential energy.
Since both masses start at rest and end at rest, we can conclude that the change in potential energy for m1 is equal to the change in potential energy for m2. In other words:
m1 * g * Δh = m2 * g * h
where m1 and m2 are the masses, g is the acceleration due to gravity, and Δh is the change in height for m1.
Now, let's substitute the given values into the equation:
(3.4 kg) * (9.8 m/s^2) * Δh = (4.1 kg) * (9.8 m/s^2) * (1.4 m)
Simplifying the equation:
33.32 kg * Δh = 56.812 kg * m
Dividing both sides of the equation by 33.32 kg:
Δh = (56.812 kg * m) / 33.32 kg
Calculating this value:
Δh ≈ 1.706 m
Therefore, m1 goes approximately 1.706 meters higher after m2 has landed.