An object with mass m1 = 6.00 kg, rests on a frictionless horizontal table and is connected to a cable that passes over a pulley and is then fastened to a hanging object with mass m2 = 8.0 kg, as shown in the figure below. Find the acceleration of each object and the tension in the cable.

There is no figure below. Kindly show your own work.

If you draw the FBD for both object separately. Then,

Positive being up and right,
T=m1a Eq 1
T-m2g=-m2a Eq 2
T= -m2a+m2g Eq 3

Eq 3 = Eq 1

m1a= -m2a+m2g
a=m2g/(m1+m2)

since, a=x
then, T=m1x

To find the acceleration of each object and the tension in the cable, we can use Newton's second law and consider the forces acting on each object.

Let's assume that the hanging object with mass m2 is accelerating downward, and the object with mass m1 is moving to the right.

1. Acceleration of the hanging object (m2):
- The only force acting on m2 is its weight (mg2), directed downward.
- Using Newton's second law: F = ma, where F is the net force and a is the acceleration.
- Net force on m2: Fnet = mg2 = m2 * g (where g is the acceleration due to gravity)
- Therefore, m2 * g = m2 * a
- Solving for a, we get a = g.

So, the hanging object with mass m2 is accelerating downward at a rate equal to the acceleration due to gravity, which is approximately 9.8 m/s^2.

2. Acceleration of the object on the table (m1):
- The force responsible for the acceleration of m1 is the tension in the cable (T).
- The weight of m1 (mg1) is equal to the normal force (since the table is frictionless).
- Using Newton's second law: F = ma, where F is the net force and a is the acceleration.
- Net force on m1: Fnet = T - mg1
- Since the only force acting on m1 is the tension T, we have T - mg1 = m1 * a
- Solving for a, we get a = (T - mg1) / m1

3. Tension in the cable:
- Since the hanging object (m2) is connected to the object on the table (m1) by the cable, the tension in the cable is the same for both objects.
- Therefore, Tension (T) = mg2 = m2 * g.

So, the acceleration of the hanging object (m2) is g (approximately 9.8 m/s^2), the acceleration of the object on the table (m1) is (T - mg1) / m1, and the tension in the cable is m2 * g.

Note: Please note that the values of masses (m1 and m2) are provided in the problem statement.

To find the acceleration of each object and the tension in the cable, we can use Newton's second law of motion.

1. Start by drawing a free-body diagram for each object:

- For object 1 (the one on the table): Since it is resting on a frictionless surface, the only force acting on it is the tension in the cable, directed upwards.
- For object 2 (the hanging object): The weight (mg) is acting downwards, and the tension in the cable is acting upwards.

2. Write down the equations of motion for each object:

- For object 1: The tension in the cable is the only force, so we have: T = m1 * a1, where T is the tension and a1 is the acceleration of object 1.
- For object 2: The weight is acting downwards, so we have: m2 * g - T = m2 * a2, where g is the acceleration due to gravity and a2 is the acceleration of object 2.

3. Eliminate the tension in the equations by substituting the first equation into the second equation:

m2 * g - m1 * a1 = m2 * a2

4. Solve the equation for acceleration:

a2 = (m2 * g - m1 * a1) / m2

5. Since the two objects are connected by a cable, their accelerations must be equal:

a1 = a2

6. Substitute the value of a1 into the equation:

a2 = (m2 * g - m1 * a1) / m2

7. Substitute values into the equation:

m2 * g - m1 * a1 = m2 * a1

Simplify the equation to solve for a1:

m2 * g = (m1 + m2) * a1

8. Solve for a1:

a1 = (m2 * g) / (m1 + m2)

9. Substitute the value of a1 into the equation to get a2:

a2 = (m2 * g) / (m1 + m2)

10. Finally, to find the tension in the cable, substitute a1 into the first equation:

T = m1 * a1

Substitute values:

T = m1 * [(m2 * g) / (m1 + m2)]

Now, you can plug in the known values (m1 = 6.00 kg, m2 = 8.0 kg, and g = 9.8 m/s^2) to find the acceleration of each object and the tension in the cable.