Two objects are connected by a light string passing over a...

Two objects are connected by a light string passing over a light, frictionless pulley as shown in Figure P8.7. The object of mass $m_{1}=5.00 \mathrm{kg}$ is released from rest at a height $h=4.00 \mathrm{m}$ above the table. Using the isolated system model, (a) determine the speed of the object of mass $m_{2}=3.00 \mathrm{kg}$ just as the 5.00 -kg object hits the table and $(\mathrm{b})$ find the maximum height above the table to which the $3.00-\mathrm{kg}$ object rises.

Two objects are connected by a light string passing over a light, frictionless pulley as shown in Figure P8.7. The object of mass $m_{1}=5.00 \mathrm{kg}$ is released from rest at a height $h=4.00 \mathrm{m}$ above the table. Using the isolated system model, (a) determine the speed of the object of
mass $m_{2}=3.00 \mathrm{kg}$ just as the 5.00 -kg object hits the table and $(\mathrm{b})$ find the maximum height above the table to which the $3.00-\mathrm{kg}$ object rises.

Key Concepts

Conservation of Mechanical Energy

This principle states that in the absence of non-conservative forces (like friction), the total mechanical energy of a system remains constant. In the context of this problem, it is used to equate the loss in gravitational potential energy with the gain in kinetic energy as the masses move, ensuring that energy is conserved throughout the motion.

Isolated System

An isolated system is one where no external forces, or net external work, act on the system. For this problem, treating the two masses connected by a pulley as an isolated system allows us to ignore external energy influences and apply energy conservation directly to the system's internal energy exchanges between potential and kinetic forms.

Pulley Systems Dynamics

Pulley systems involve connecting masses via a string that passes over a pulley, where the motion of one mass affects the other. Even though the masses may experience different accelerations due to gravity, the connection by the string ensures that energy and momentum are transferred between them in a predictable manner. Understanding the role of the pulley, especially assuming it is frictionless and the string is light, simplifies the analysis by removing rotational dynamics and focusing on translational motion.

Gravitational Potential Energy and Kinetic Energy Conversion

This concept focuses on the transformation of energy from gravitational potential energy to kinetic energy as objects move in a gravitational field. As one object descends and loses height, its potential energy decreases, and correspondingly its kinetic energy increases, affecting the motion of both objects connected in the system. This conversion is key to solving for velocities and heights reached during the motion.

Transcript

00:01 For this problem on the topic of conservation of energy, we are given two objects which are connected by a light string that passes over a light frictionless pulley as shown in the figure.

00:10 We know the mass of object one, and we know the point from which it is released above the table.

00:18 Now, using the isolated system model, we are asked to determine the speed of the object of mass m2, the mass of which is given, as the first object hits the table, and then to find the maximum height above the table, to which the second mass rises.

00:36 Now we'll assign the height y is equal to zero to the table top and then use the conservation of energy for the system of the earth and the two objects.

00:46 If we choose the initial point before release and the final point which we will code as with subscript faa just before the larger object hits the floor, we know that no external forces do work on the system and no friction acts on the system.

01:00 The total mechanical energy remains constant, and we can write the energy version of the isolated system model as follows.

01:09 So the kinetic energy at a plus the kinetic energy at point b plus the gravitational potential energy ug initially must equal to the kinetic energy of mass a plus the kinetic energy of mass b plus the gravitational potential energy.

01:37 Finally which is f -a now at the initial point k -a -i and k -b -i are zero we have no kinetic energy initially so we define the gravitational potential energy of the system as zero and so the total initial energy is zero and we have that zero is equal to a half into m1 plus m2 into vfa squared plus m2 times gh plus m1 times g minus h.

02:31 Here we've used the fact that because the core does not stretch, the two blocks have the same speed.

02:37 The heavier mass moves down, losing gravitational potential energy as the lighter mass moves up and gains gravitational potential energy...

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