Two objects with masses 5.00 kg and 2.00 kg hang 0.600 m above the floor from the ends of a cord

Two objects with masses 5.00 kg and 2.00 kg hang 0.600 m...

Key Concepts

Conservation of Mechanical Energy

Energy conservation is a powerful tool in analyzing the system’s behavior over different stages of motion. It relates the transformation of gravitational potential energy into kinetic energy during the upward or downward motion and is particularly useful when determining additional travel after a system's constraint is removed, like after a mass hits the floor.

Transition from Constrained Motion to Projectile Motion

In some setups, one part of the system may reach a physical limit (for example, an object hitting the floor), which changes the constraints on the system. At that instant, the other mass may continue moving upward as a free projectile under gravity; hence, understanding how to treat the motion both under the influence of a constraint and as a separate projectile is important.

Atwood Machine

An Atwood machine is a system consisting of two masses connected by a cord over a pulley, often idealized as frictionless and massless. This setup illustrates how a difference in gravitational forces on the two masses produces an acceleration in the system, making it a classic example in dynamics for analyzing tension and acceleration.

Newton's Second Law for Connected Bodies

In problems involving connected bodies, Newton's second law is applied to each mass to set up equations of motion. The net force acting on each mass is determined by the difference between gravitational forces and the tension in the cord, allowing one to solve for the acceleration of the system, which is key when the system is released from rest.

Kinematics of Uniform Acceleration

Once the acceleration is determined, kinematics comes into play to analyze the motion over time. Kinematic equations for constant acceleration help in calculating displacements, velocities, and times during the motion of the masses, such as the distance a rising object travels before reaching its maximum height.

Transcript

00:01 This problem is a simple at -woods machine, and so we can refer to the discussion in the text where they derive the acceleration of either mass in an out -woods machine, and they get acceleration is g, m -1 minus m -2, over m -1 plus m -2.

00:22 If you're confused on where this came from, it's a simple application of newton's second law in the y direction, and then you eliminate the tension to solve for the acceleration, and you get this.

00:33 But we'll go ahead and start with this since it's such a famous expression in the outwoods machine.

00:39 And so plugging it in the masses they give us, we determine that the acceleration of the 2 -kilogram object is equal to 3 -7s -g, and it will go upwards, whereas the acceleration of the 5 -kilogram object is the same magnitude, but it'll go downward since it's heavier.

01:06 Now that we know that, we're going to let the initial height, we're going to let the initial height where both these masses start off be equal to h0.

01:22 And so now this becomes a kinematic equation problem now that we've determined the acceleration.

01:29 So when the 5 kilogram mass hits the ground, the 2 kilogram mass is at a height 2h0.

01:36 And so the starting distance is actually 2h0.

01:42 But we want to figure out what the velocity of the 2 kilogram object is when it reaches the height to h0 when the 5 kilogram mass hits the ground.

01:53 And so the distance it travels from 2h0 back to h0 is just h0.

02:02 The acceleration we found is 3 .7s g.

02:06 The initial velocity of the 2 kilogram object is zero.

02:09 And the final velocity after the 5 kilogram object is what is desired.

02:15 So for this, we're going to use this kind of mac equation and solve for v.

02:25 V, not zero, so that turn goes away...

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