In the Atwood machine shown below, m1 = 2.00 kg and m2 = 6.00 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 = 2.60 m/s downward.
(a) How far will m1 descend below its initial level?
(b) Find the velocity of m1 after 1.80 s.
To solve this problem, we can use the principles of Newton's second law and the concepts of conservation of energy.
(a) To determine how far m1 will descend below its initial level, we first need to find its acceleration. In an Atwood machine, the acceleration can be calculated using the formula:
a = (m2 - m1) * g / (m1 + m2)
where m1 and m2 are the masses of the objects, and g is the acceleration due to gravity (9.8 m/s^2).
Substituting the given values:
m1 = 2.00 kg
m2 = 6.00 kg
a = (6.00 kg - 2.00 kg) * 9.8 m/s^2 / (2.00 kg + 6.00 kg)
Simplifying this expression gives us:
a = 3.92 m/s^2
Now that we know the acceleration, we can calculate the distance m1 will descend. We can use the following kinematic equation:
d = vi * t + 0.5 * a * t^2
where vi is the initial velocity, t is the time, and d is the distance.
Substituting the given values:
vi = 2.60 m/s (downward)
t = ?
a = 3.92 m/s^2
d = ?
We want to find the distance m1 will descend, so we need to solve for d. Rearranging the equation, we get:
d = vi * t + 0.5 * a * t^2
Since m1 starts from rest, its initial velocity is 0. Therefore, the equation simplifies to:
d = 0 + 0.5 * a * t^2
Now, we can substitute the given time:
t = 0.00 s
Substituting these values into the equation, we get:
d = 0 + 0.5 * 3.92 m/s^2 * (0.00 s)^2
Simplifying, we find that the distance m1 descends is:
d = 0
Therefore, m1 does not descend below its initial level.
(b) To find the velocity of m1 after 1.80 s, we can use the same kinematic equation as before:
d = vi * t + 0.5 * a * t^2
but this time we want to solve for vi, the final velocity of m1.
Rearranging the equation, we get:
vi = (d - 0.5 * a * t^2) / t
Substituting the given values:
d = ?
a = 3.92 m/s^2
t = 1.80 s
We know that m1 does not descend from its initial level, so d = 0. Therefore, the equation becomes:
vi = (0 - 0.5 * 3.92 m/s^2 * (1.80 s)^2) / 1.80 s
Simplifying, we find:
vi = -6.998 m/s (downward)
Therefore, the velocity of m1 after 1.80 s is approximately -6.998 m/s downward.