A block of mass m1 = 6.25 kg on a frictionless plane inclined at angle θ = 30.7° is connected by a cord over a massless, frictionless pulley to a second block of mass m2 = 2.36 kg hanging vertically (Fig. 5-52). (a) What is the acceleration of the hanging block (choose the positive direction up)? (b) What is the tension in the cord?

Question

A block of mass m1 = 6.25 kg on a frictionless plane inclined at angle θ = 30.7° is connected by a cord over a massless, frictionless pulley to a second block of mass m2 = 2.36 kg hanging vertically (Fig. 5-52). (a) What is the acceleration of the hanging block (choose the positive direction up)? (b) What is the tension in the cord?

Answer 1

To find the acceleration of the hanging block and the tension in the cord, we can use Newton's second law of motion.

First, let's analyze the forces acting on each block separately:

For the block on the inclined plane (m1):
1. There is the force of gravity pulling downward with magnitude m1 * g, where g is the acceleration due to gravity (approximately 9.8 m/s²).
2. There is a component of the gravitational force acting parallel to the inclined plane, which is m1 * g * sin(θ), where θ is the angle of inclination.
3. There is also a normal force acting perpendicular to the inclined plane, which balances the component of the gravitational force perpendicular to the plane. It has the same magnitude but opposite direction as m1 * g * cos(θ).

For the hanging block (m2):
1. There is the force of gravity pulling downward with magnitude m2 * g.

Now, let's determine the relationship between the acceleration of the blocks and the tension in the cord. Since the blocks are connected by a cord over a pulley, they will have the same magnitude of acceleration, denoted as "a". The hanging block (m2) will have an upward acceleration, while the block on the inclined plane (m1) will have a downward acceleration.

With these forces in mind, let's apply Newton's second law to each block:

For m1:
Summing the forces parallel to the inclined plane, we have:
m1 * a = m1 * g * sin(θ) - T
(where T is the tension in the cord)

For m2:
Summing the forces acting along the vertical direction, we have:
m2 * g - T = m2 * a

Now, we have two equations with two unknowns (a and T). We can solve them simultaneously to get the values.

Let's substitute the given values into the equations:
m1 = 6.25 kg, m2 = 2.36 kg, θ = 30.7°, g = 9.8 m/s².

For the equation of m1:
6.25 kg * a = 6.25 kg * (9.8 m/s²) * sin(30.7°) - T

For the equation of m2:
2.36 kg * (9.8 m/s²) - T = 2.36 kg * a

Now, solve these equations simultaneously to find the values of "a" and "T".