# Video: Finding the Work Done by a Ring Sliding down a Vertical Pole

A ring of mass 1.5 kg was sliding down a vertical pole. Starting from rest, it accelerated over a distance of 3.3 m until its speed became 6.2 m/s. Using the work-energy principle, determine the work done by the resistance to the ring’s motion. Take 𝑔 = 9.8 m/s².

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### Video Transcript

A ring of mass 1.5 kilograms was sliding down a vertical pole. Starting from rest, it accelerated over a distance of 3.3 meters until its speed became 6.2 meters per second. Using the work energy principle, determine the work done by the resistance to the ring’s motion. Take 𝑔 equals 9.8 meters per second squared.

Let’s start by recording some of the helpful information given in this statement. We’re told the mass of the ring is 1.5 kilograms. We’ll name that 𝑚. The ring travels a distance of 3.3 meters down a pole. We’ll call that distance 𝑑. After traveling that distance, it’s attained a speed of 6.2 meters per second, what we’ll call 𝑣 sub 𝑓. We’re told further that the acceleration due to gravity 𝑔 is equal to 9.8 meters per second squared. And we want to solve for the work done by the resistance to the ring’s motion. We’ll call that work 𝑊 sub 𝑟.

Let’s draw a diagram of this situation. We have a ring, whose mass is known, which slides a particular distance 𝑑, also known, down a vertical pole, achieving a velocity we’ve called 𝑣 sub 𝑓.

If we were to draw in the forces that the ring experiences, there’s the gravitational force that draws the ring down, 𝑚 times 𝑔. There’s also a force that acts upward that resists the descent of the ring on the pole. We can call that 𝐹 sub 𝑟 for the resistive force. This force is due to friction.

We want to solve for the work done by this resistive force. And to do it, we’re gonna recall a principle called the work energy principle. This principle says that the work done on an object 𝑊 is equal to its change in kinetic energy. So we could write that the work done by the resistive force is equal to the change in kinetic energy explicitly caused by that resistance.

To write out an expression for the change in kinetic energy due to the resistive force, we need to keep two things in mind. First, whether there’s a resistive force or not, our ring would gain kinetic energy due to the force of gravity pulling it down the pole. And second, the values we’ve been given for the final speed of the ring and the distance that it traveled reflect the resistance that happens as it slides.

We could write that the change in kinetic energy explicitly due to resistance is equal to the energy of the ring at the top of its path before it starts moving minus its real energy at the bottom of its path, that is, after it’s descended 3.3 meters.

If we recall that an object’s gravitational potential energy is equal to its mass times 𝑔 times its height above some baseline and that an object’s kinetic energy is equal to half its mass times its speed squared, then we can replace 𝐸 sub top with 𝑚 times 𝑔 times 𝑑. And we can replace 𝐸 sub bottom with one-half 𝑚 𝑣 sub 𝑓 squared.

Looking at these two terms, if there is no resistive force on the ring as it slid down, then 𝑚𝑔𝑑 minus one-half 𝑚 𝑣 sub 𝑓 squared would be zero. That is, all of the potential energy would be perfectly converted to kinetic energy. But because some of the energy of the ring is lost to friction, these two terms will not be exactly equal. And their difference is equal to the energy lost due to that resistance.

Since we’re given the mass of the ring as well as the distance it descends, its final speed, and the acceleration due to gravity, we’re ready to plug in and solve. When we do, having factored out the mass, and enter these terms on our calculator, we find that the change in kinetic energy due to resistance is 19.68 joules. And by the work energy principle, this is equal to the work done by the resistive force.