Video: Applying the Conversion of Kinetic Energy to Gravitational Potential Energy to Determine a Required Velocity

Ignoring details associated with friction, extra forces exerted by arm and leg muscles, and other factors, we can consider a pole vault as the conversion of an athlete’s running kinetic energy to gravitational potential energy. If an athlete is to lift his body 4.8 m during a vault, what speed must he have when he plants his pole?

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Video Transcript

Ignoring details associated with friction, extra forces exerted by arm and leg muscles, and other factors, we can consider a pole vault as the conversion of an athlete’s running kinetic energy to gravitational potential energy. If an athlete is to lift his body 4.8 meters during a vault, what speed must he have when he plants his pole?

As we solve this problem along with the assumption that the other factors mentioned at the top of the problem statement are negligible, we’ll also assume that the acceleration due to gravity, 𝑔, is exactly 9.8 metres per second squared. And further, even though we may draw a sketch of our pole vaulter as an extended body, we’ll treat him mathematically as a single point when we do our calculations.

So those are our assumptions. And again what we’re looking to do is to solve for a speed, we can call it 𝑣, that the athlete must obtain in order to lift his body 4.8 meters above ground during the pole vault. So let’s draw a simple sketch of what’s going on.

So here we have our athlete holding a pole for the pole vault and running towards this set of uprights where the pole is positioned 4.8 meters above the ground. Now the athlete must be running at a speed 𝑣 in order to achieve that height, and that’s what we want to solve for.

Now we’re talking here about two different forms of energy: kinetic and potential. And let’s recall that these forms altogether are conserved throughout a process. That’s called the conservation of energy. That law tells us that energy cannot be created or destroyed. It can only be converted from one form to another.

In our case, that means that the kinetic energy of the athlete will be converted entirely to potential energy when he is at its maximum height. There is no loss of energy, but rather it’s simply converted from one form to another.

We can write that as one half 𝑚𝑣 squared where 𝑚 is the mass of the athlete and 𝑣 is his speed is equal to 𝑚𝑔ℎ: the mass of the athlete times 𝑔, the acceleration due to gravity, times ℎ, the height the athlete obtains.

This is another way of saying that the athlete’s kinetic energy is entirely converted into gravitational potential energy. As we look at this equation, we see that 𝑣 is the variable we want to solve for. That’s the speed the athlete must obtain in order for ℎ, the height, to be 4.8 metres.

One thing to notice about this equation is that the athlete’s mass is on both sides, which means that it cancels out of the equation. Now that’s interesting, because it says that in this process it has nothing to do with the mass of the athlete. It could be a large or small number, and our result is independent of it. So we are left with a simplified equation which reads that one half the speed of the athlete squared is equal to the acceleration due to gravity times ℎ, where ℎ, as we’ve been given in the problem, is 4.8 metres.

To rearrange this and solve for 𝑣, we’ll first multiply both sides by two. The two and one half on the left side of our equation cancel out. And if as a next step we take the square root of both sides, then that square root and the square over the 𝑣 cancel out, leaving us with an expression that reads 𝑣, the speed of the athlete, is equal to the square root of two times the acceleration due to gravity times ℎ.

We can plug in the values we know for 𝑔 and ℎ. With those values inserted, we type this expression into our calculator and find a speed 𝑣 that the athlete must obtain of 9.7 metres per second. That’s the speed the athlete must run at in order to reach a height of 4.8 metres above the ground.

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