In an Atwood machine m1 = 2.00 kg and m2 = 6.50 kg. The masses of the pulley and string are negligible by comparison. The pulley turns without friction and the string does not stretch. The lighter object is released with a sharp push that sets it into motion at vi = 2.20 m/s downward.
(a) How far will m1 descend below its initial level?
___________m
(b) Find the velocity of m1 after 1.80 ____________s.
Please explain this as thoroughly as possible! thanks!
I assume m2 is the hanging mass.
the pulling force is m2*g
the pulling masses are M1+m2
Forcepulling=totalmassesmoving*a
now using a,
solve for distance and velocity.
To solve this problem, we can use the concept of conservation of energy. The total mechanical energy of the system is conserved, which means that the sum of the potential energy and kinetic energy remains constant.
(a) To find how far m1 will descend below its initial level, we need to determine the change in potential energy.
The potential energy (PE) of an object of mass m at a height h above its reference level can be given by:
PE = m * g * h,
where g is the acceleration due to gravity (9.8 m/s²).
Initially, m1 is at a height h above the reference level. At the final position, m1 descends by a distance y. The potential energy at the initial position is given by:
PE_initial = m1 * g * h.
At the final position, the potential energy is given by:
PE_final = m1 * g * (h + y).
Since the total mechanical energy is conserved, we can equate the initial and final potential energies:
PE_initial = PE_final.
m1 * g * h = m1 * g * (h + y).
Canceling out the g and m1 terms, we get:
h = h + y.
Simplifying the equation, we find:
y = 0.
This means that m1 will not descend below its initial level. It will remain at the same height.
(b) To find the velocity of m1 after 1.80 s, we can use the concept of acceleration.
The net force acting on m1 is the difference between the gravitational force and the tension force.
Net force = m1 * g - T,
where T is the tension in the string.
Using Newton's second law of motion, F = ma, we can express the acceleration of m1 as:
m1 * a = m1 * g - T.
Since m1 and T are equal in magnitude and opposite in direction, we can rewrite the equation as:
m1 * a = m1 * g - m1 * g2,
where g2 is the acceleration due to gravity acting on m2.
Simplifying the equation, we find:
a = g - g2.
Using the kinematic equation, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can find the final velocity of m1 after 1.80 s:
v = vi + a * t.
Substituting the given values:
v = 2.20 m/s + (9.8 m/s² - 6.5 m/s²) * 1.80 s.
Simplifying the equation, we find:
v = 2.20 m/s + 3.3 m/s² * 1.80 s.
Calculating the value, we find:
v ≈ 7.04 m/s.
Therefore, the velocity of m1 after 1.80 s is approximately 7.04 m/s.
To solve this problem, we can use the principles of Newton's second law and the conservation of energy. Let's break down the problem into two parts: finding the distance m1 descends below its initial level and finding the velocity of m1 after a given time.
(a) How far will m1 descend below its initial level?
To calculate the distance m1 descends, we need to find the acceleration of the system first. In an Atwood machine, the net force on the system is given by the difference in the weights of the two masses. The formula for finding the acceleration (a) in an Atwood machine is:
a = (m2 - m1) * g / (m1 + m2)
where m1 is the mass of object 1 (2.00 kg), m2 is the mass of object 2 (6.50 kg), and g is the acceleration due to gravity (9.8 m/s^2).
Plugging in the values, we get:
a = (6.50 kg - 2.00 kg) * 9.8 m/s^2 / (2.00 kg + 6.50 kg)
a = 4.50 kg * 9.8 m/s^2 / 8.50 kg
a ≈ 5.235 m/s^2
Now, we can use the kinematic equation to find the distance (d) m1 descends. The equation is:
d = v^2 - vi^2 / (2a)
where v is the final velocity, vi is the initial velocity, and a is the acceleration.
Since m1 is released with a sharp push and we want to find how far it descends below its initial level, the final velocity (v) will be zero. Hence, we have:
d = 0 - (2.20 m/s)^2 / (2 * 5.235 m/s^2)
d = -4.84 / 10.47
d ≈ -0.462 m
The negative sign indicates that m1 descends below its initial level. However, the magnitude of the displacement is positive, so the value is approximately 0.462 meters.
(b) Find the velocity of m1 after 1.80 seconds.
To find the velocity of m1 at a given time, we can use the equation:
v = vi + at
where v is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.
Using the values, we have:
v = 2.20 m/s + 5.235 m/s^2 * 1.80 s
v ≈ 11.25 m/s
So, the velocity of m1 after 1.80 seconds is approximately 11.25 m/s.