A block of mass m=5.00 kg is released from point @ and...

A block of mass $m=5.00 \mathrm{~kg}$ is released from point $@$ and slides on the frictionless track shown in Figure $\mathrm{P} 8.3 .$ Determine (a) the block's speed at points (B) and @ and (b) the net work done by the gravitational force on the block as it moves from point $(\mathrm{A}$ to point $\mathrm{C}$ ).

Key Concepts

Kinetic Energy

Kinetic energy is the energy an object has due to its motion, defined as (1/2)mv². By using the known initial potential energy and applying conservation principles, one can calculate the speed of an object at different points along a frictionless path based on the variation in kinetic energy.

Gravitational Potential Energy

Gravitational potential energy is the energy an object possesses due to its position in a gravitational field, typically expressed as mgh, where m is mass, g is the acceleration due to gravity, and h is the height relative to a reference point. Its conversion to kinetic energy during motion is a crucial aspect of energy analysis in mechanics.

Conservation of Mechanical Energy

This principle states that in the absence of non-conservative forces (like friction), the total mechanical energy of a system (the sum of its potential and kinetic energies) remains constant. This allows for the direct conversion of gravitational potential energy into kinetic energy as the object moves, which is key to determining its speed at various points along a frictionless path.

Work-Energy Theorem

The work-energy theorem links the net work done on an object to its change in kinetic energy. In systems where non-conservative forces are absent or accounted for separately, the work done by gravity (a conservative force) directly corresponds to the change in an object’s kinetic energy as it moves along its path.

Transcript

00:01 Here we have an energy conservation problem.

00:03 We have a block with mass 5 kilograms that we start out here at point a, and then we drop it down, it drops down to point b and point c.

00:13 For the first part of the problem, we want to know what is the speed of the block at points b and c.

00:18 So we know the block has no velocity at point a, since that's where we release it.

00:25 And so if we want to use energy conservation, we should just find the initial energy.

00:32 So the initial energy is going to be all potential energy, ultimately.

00:38 It's going to be equal to mgh, where this h is going to be five.

00:43 So let's maybe write this out as just five meters, to be more explicit.

00:49 So that is the energy initial of the block.

00:53 Now at point b, this is going to have a potential energy of 3 .2 meters.

01:03 And then, of course, it's also going to have some velocity.

01:07 So 1 .5m2 squared.

01:10 And so now we have here an equation in which we can solve for the velocity at point b here.

01:19 So let's just solve for that, right? let's see.

01:24 So first, let's just move over this mg term.

01:26 So 1 1 .5mvb squared is going to be equal to mg.

01:34 Subtracting that over it'll be 1 .8 meters it looks like the masses cancelled out and vb is going to be equal to the square root of 2g 1 .8 meters all right and what this velocity is is 5 .94 meters per second okay now similarly you can do the same thing for point c...

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