a mass m2 of 1 kg is plqaced on a frictionless inclined plane inclined at angle of 30 degrees.it is connected by a chord of negligble mass to a mass m1 of 3 kg on horizontal frictionless table. if a force of magnitude 2.3n is applied on mass m1 towards right then what is the tension in chord ?

To find the tension in the chord connecting the two masses, we need to analyze the forces acting on both masses separately. Let's break it down step by step.

Understanding the System

We have two masses: mass m2 (1 kg) on a frictionless inclined plane at an angle of 30 degrees, and mass m1 (3 kg) on a horizontal frictionless table. A force of 2.3 N is applied to m1 towards the right. The tension in the chord will be influenced by both the gravitational force acting on m2 and the applied force on m1.

Forces Acting on m2

For mass m2, the forces acting on it are:

• The gravitational force acting downwards: F_gravity = m2 * g
• The component of gravitational force acting parallel to the incline: F_parallel = m2 * g * sin(θ)

Here, g is the acceleration due to gravity (approximately 9.81 m/s²), and θ is the angle of the incline (30 degrees).

Calculating the parallel component:

F_parallel = 1 kg * 9.81 m/s² * sin(30°)

Since sin(30°) = 0.5, we have:

F_parallel = 1 kg * 9.81 m/s² * 0.5 = 4.905 N

Forces Acting on m1

For mass m1, the forces acting on it are:

• The applied force of 2.3 N to the right.
• The tension T in the chord acting to the left.

Using Newton's second law (F = ma), we can express the net force acting on m1:

Net Force on m1 = Applied Force - Tension

Thus, we have:

2.3 N - T = m1 * a

Where a is the acceleration of the system.

Finding the Acceleration

Now, we need to find the acceleration of the system. The acceleration of m2 down the incline is equal to the acceleration of m1 on the table since they are connected by the chord. For m2, we can write:

Net Force on m2 = T - F_parallel

Using Newton's second law again:

T - 4.905 N = m2 * a

Substituting m2 = 1 kg:

T - 4.905 N = 1 kg * a

Now we have two equations:

1. 2.3 N - T = 3 kg * a
2. T - 4.905 N = 1 kg * a

Solving the Equations

From the second equation, we can express T in terms of a:

T = 4.905 N + a

Substituting this expression for T into the first equation:

2.3 N - (4.905 N + a) = 3 kg * a

Rearranging gives:

2.3 N - 4.905 N = 3 kg * a + a

-2.605 N = 4 kg * a

Thus, we find:

a = -2.605 N / 4 kg = -0.65125 m/s²

Finding the Tension

Now that we have the acceleration, we can substitute it back into the equation for T:

T = 4.905 N + (-0.65125 m/s²)

T = 4.905 N - 0.65125 N = 4.25375 N

Therefore, the tension in the chord is approximately 4.25 N.