An Atwood's machine consists of masses m1 and m2, and a pulley of negligible mass and friction. Starting from rest, the speed of the two masses is 3.80 m/s at the end of 3.16 s. At that time, the kinetic energy of the system is 77.0 J and each mass has moved a distance of 6.00 m. Determine the lighter mass.

To determine the lighter mass in this Atwood's machine, we can use the concept of conservation of energy.

The total kinetic energy of the system is given as 77.0 J. Since there is no external force or friction acting on the system, the total mechanical energy of the system is conserved.

The kinetic energy, K, can be calculated using the formula K = (1/2) * m * v^2, where m represents the mass of an object and v represents its velocity.

Let's denote the mass m1 as the lighter mass, and m2 as the heavier mass.

Given:
Speed of masses at the end of 3.16 s, v = 3.80 m/s
Distance moved by each mass, d = 6.00 m
Kinetic energy of the system, K = 77.0 J

We can relate the speed of the masses to their distance traveled and their time using the formula v = d / t. Rearranging this equation gives us d = v * t.

Considering the conservation of energy:
The total kinetic energy of the system can be expressed as:
K = (1/2) * m1 * v1^2 + (1/2) * m2 * v2^2

Since both masses have the same displacement, d, we can write v1 = -v2, where the negative sign indicates that the lighter mass is moving up while the heavier mass is moving down.

Substituting v1 = -v2 into the expression for kinetic energy, we have:
K = (1/2) * m1 * (-v2)^2 + (1/2) * m2 * v2^2

Since v1 = -v2, we can rewrite the expression above as:
K = (1/2) * (m1 + m2) * v2^2

Substituting the given values into the equation, we have:
77.0 J = (1/2) * (m1 + m2) * (3.80 m/s)^2

Simplifying further, we get:
77.0 J = (0.5) * (m1 + m2) * 14.44 m^2/s^2
154 J = (m1 + m2) * 14.44 m^2/s^2
m1 + m2 = 154 J * (s^2/m^2) / 14.44 m^2/s^2
m1 + m2 = 10.6775 kg

We also know that the distance traveled by each mass is 6.00 m. Using the formula d = v * t, we can find the time taken, t:
6.00 m = 3.80 m/s * t
t = 6.00 m / (3.80 m/s)
t = 1.579 s

Since the masses started from rest, the acceleration, a, can be calculated using the equation v = u + at, where u is the initial velocity. Substituting the values, we have:
3.80 m/s = 0 + a * 1.579 s
a = 3.80 m/s / 1.579 s
a = 2.408 m/s^2

Now, we can use the equation of motion, x = ut + (1/2)at^2, to find the displacement of a mass.
Since the masses move equal distances, we can choose any of them. Let's choose m1:
6.00 m = 0 + (1/2) * 2.408 m/s^2 * (1.579 s)^2
6.00 m = 0 + 1.891 m
m = 6.00 m / 1.891 m
m = 3.17 kg

Now, we know that m1 + m2 = 10.6775 kg. Substituting the value of m2 as 3.17 kg, we can solve for m1:
m1 = 10.6775 kg - 3.17 kg
m1 = 7.5075 kg

Therefore, the lighter mass, m1, is approximately 7.51 kg.