The device called an At-woods machine.if m2 is greatr than m1 an assumng th pulley to be frictionless and massless.1.how far will m2 fall in th time T aftr the system is released? 2.what is th tension in the cord that connect two massess? 3. If m1=5kg and m2=3kg,what is the tension in the cord? 4. What is the time needed for the masses to move 15m from rest?

To answer your questions about the Atwood's machine, we can use the principles of Newton's laws of motion, particularly the concept of force and acceleration. Here are the explanations for each question:

1. How far will m2 fall in the time T after the system is released?
To find the distance m2 falls, we need to determine the acceleration of the system. The net force on the system is given by the difference in tension forces on either side of the pulley. Assuming the pulley is frictionless and massless, the tension on one side of the pulley is greater than the other by the weight difference between the two masses, which is equal to (m2 - m1) * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Using Newton's second law, which states that force equals mass times acceleration (F = ma), we can write the equation for the system as (m2 - m1) * g = (m2 + m1) * a, where a is the acceleration of the system.

From here, we can rearrange the equation to solve for acceleration (a): a = (m2 - m1) * g / (m2 + m1).

Using the kinematic equation, s = ut + (1/2) * a * t^2, where s is the distance fallen, u is the initial velocity (which is zero in this case), and t is the time, we can substitute the values to find the distance m2 falls (s):
s = 0 + (1/2) * [(m2 - m1) * g / (m2 + m1)] * t^2.
Simply plug in the given values and solve for s using the desired time T.

2. What is the tension in the cord that connects the two masses?
The tension in the cord can be determined by considering the forces acting on each mass. For mass m1, the tension force T1 in the cord acts upward, balancing the weight force m1 * g acting downward. For mass m2, the tension force T2 in the cord acts downward, balancing the weight force m2 * g acting upward.

The tension in the cord is the same on both sides since it is a single continuous cord. Therefore, T1 = T2.

3. If m1 = 5 kg and m2 = 3 kg, what is the tension in the cord?
Given the masses, we can calculate the tension using the equation derived in response to the previous question. By substituting the given values into the equation T1 = T2, we can find the tension in the cord.

4. What is the time needed for the masses to move 15 m from rest?
To determine the time needed for both masses to move a distance of 15 m from rest, we can use the kinematic equation s = ut + (1/2) * a * t^2.

Since the initial velocity, u, is zero, the equation simplifies to s = (1/2) * a * t^2.

Knowing the acceleration from the previous explanation, we can substitute the given distance (s = 15 m) and solve for t, the time required for the masses to move the specified distance.