A light, inextensible cord passes over a light,

frictionless pulley with a radius of 11 cm. It
has a(n) 14 kg mass on the left and a(n)
4.3 kg mass on the right, both hanging freely.
Initially their center of masses are a vertical
distance 2.1 m apart.
The acceleration of gravity is 9.8 m/s2 .At what rate are the two masses accelerat-
ing when they pass each other?
Answer in units of m/s2
What is the tension in the cord when they
pass each other?
Answer in units of N

EXPLAIN HOW you got your anwser?

To find the rate at which the two masses are accelerating when they pass each other and the tension in the cord at that moment, we can apply Newton's second law of motion for each mass.

First, let's consider the 14 kg mass on the left. The gravitational force acting on it can be calculated as the product of its mass and the acceleration due to gravity (9.8 m/s²): F_gravity_left = 14 kg × 9.8 m/s² = 137.2 N (downward).

Next, we need to find the tension in the cord. Since the cord is inextensible, the tension in it will be the same on both sides of the pulley. Let's call the tension T.

For the 4.3 kg mass on the right, the gravitational force acting on it is F_gravity_right = 4.3 kg × 9.8 m/s² = 42.14 N (downward).

Since the masses are connected by the cord, the net force on both masses is equal. The net force on the left side is given by F_net_left = T - F_gravity_left, and on the right side, it is given by F_net_right = F_gravity_right - T.

Now, we can set up an equation based on the fact that the net force is equal to the product of the mass and acceleration for each side:

For the left side: F_net_left = m_left × a
For the right side: F_net_right = m_right × a

Substituting the expressions for net force and gravitational force, we have:
T - F_gravity_left = m_left × a
F_gravity_right - T = m_right × a

Simplifying the equations:
T - 137.2 N = 14 kg × a
42.14 N - T = 4.3 kg × a

Now, we can solve this system of equations to find the acceleration (a) and the tension (T).

From the first equation:
T = 14 kg × a + 137.2 N

Substituting T in the second equation:
42.14 N - (14 kg × a + 137.2 N) = 4.3 kg × a

Simplifying the equation:
42.14 N - 137.2 N = (4.3 kg + 14 kg) × a
-95.06 N = 18.3 kg × a
a = -95.06 N / 18.3 kg ≈ -5.19 m/s²

Since the acceleration comes out to be negative, it means that the masses are accelerating towards each other.

To find the tension when the masses pass each other, we can substitute the value of a into the T equation:
T = 14 kg × (-5.19 m/s²) + 137.2 N
T ≈ 206.86 N

Therefore, the rate at which the two masses are accelerating when they pass each other is approximately 5.19 m/s² (towards each other), and the tension in the cord at that moment is approximately 206.86 N.

To find the rate at which the two masses are accelerating when they pass each other, we can use the principle of conservation of mechanical energy.

First, let's determine the masses' initial potential energy:

Mass 1 (14 kg):
Initial potential energy = m1 * g * h
= 14 kg * 9.8 m/s^2 * 2.1 m

Mass 2 (4.3 kg):
Initial potential energy = m2 * g * h
= 4.3 kg * 9.8 m/s^2 * (2.1 m + 0.22 m) // accounting for the radius of the pulley

Since the system starts from rest and there is no initial kinetic energy, we can equate the initial potential energy to the final potential energy when the masses pass each other:

m1 * g * h = m2 * g * (2.1 m - d)

Here, "d" represents the vertical distance the masses will travel before they pass each other.

Now, let's differentiate this equation with respect to time to find the rate of change of the vertical distance:

m1 * g * (dh/dt) = -m2 * g * (dd/dt)

Since the acceleration of both masses is the same, we can write:

m1 * (dh/dt) = -m2 * (dd/dt)

Substituting the given values:

14 kg * (dh/dt) = -4.3 kg * (dd/dt)

Simplifying:

(dd/dt) = -(14 kg / 4.3 kg) * (dh/dt)

The negative sign indicates that the masses are accelerating towards each other. Now, let's find the values of (dh/dt) and substitute them into the equation:

To find (dh/dt), we can relate it to the angular velocity of the pulley. Since there is no slipping at the contact point between the cord and the pulley, we have:

v = r * (dθ/dt)

Considering the circular motion of the pulley, v is the velocity of the masses and dθ/dt is their rate of change of angular position. The velocity of both masses is the same because they are connected by the cord, so we can write:

(dh/dt) = (dθ/dt) * r

Now, we need to relate (dθ/dt) to (dd/dt) by considering the difference in radii of the two masses.

When the masses pass each other, the distance traveled by mass 1 is equal to the perimeter of the pulley minus the distance traveled by mass 2:

2πr - (2π(r - h)) = 2πh

Differentiating this equation with respect to time:

0 - (2π(dh/dt)) = 2π(dh/dt)

(2π(dh/dt)) = -(2π(dd/dt))

Now, substituting (dh/dt) and (2π(dh/dt)) into the equation:

(dd/dt) = -(14 kg / 4.3 kg) * [(dh/dt) / (2πr)] * (2πr)
= -(14 kg / 4.3 kg) * (dh/dt)

Therefore, the rate at which the masses are accelerating when they pass each other is given by:

(dd/dt) = -(14 kg / 4.3 kg) * [(dh/dt) / (2πr)] * (2πr)
= -(14 kg / 4.3 kg) * (dh/dt)
= -(14 kg / 4.3 kg) * (dd/dt)

To find the tension in the cord when the masses pass each other, we can consider the forces acting on each mass.

For mass 1: T - m1 * g = m1 * (dd/dt)
For mass 2: m2 * g - T = m2 * (dd/dt)

Solving these equations simultaneously, we can find the value of T.

Let's calculate the numerical values to get the final answers.

dqw