Lesson Video: Mechanical Energy and Dynamics

In this video we learn that mechanical energy combines kinetic and potential energy, and that dynamics involves motion due to unbalanced forces.


Video Transcript

In this video, we’re going to talk about mechanical energy and dynamics. We’ll learn what mechanical energy is and we’ll see how it interacts with objects in motion due to unbalanced forces.

To get started, imagine that you are a championship-level skateboarder competing in your favorite event — the half-pipe. In the half-pipe, by rolling back and forth on a u-shaped surface, you aimed to gain enough energy so that by the time you reach the top of the other side you’re able to be airborne and perform tricks. As you move back and forth, you constantly convert potential energy to kinetic energy and back again, while also experiencing energy loss due to friction between your wheels and the surface.

To help understand your motion on the half-pipe, we want to know something about mechanical energy and dynamics. Mechanical energy is the combination of an object’s kinetic and potential energy — in other words, the addition of its energy due to motion and its potential motion. As an equation, we can write that mechanical energy equals kinetic energy plus potential energy.

We can introduce this idea of mechanical energy into the world of dynamics, which describes motion caused by forces of an unbalanced forces. We’ll see that when the system we’re considering is isolated — that is no energy is added to it or taken from it — even in the presence of forces energy is conserved and we’ll also see that Newton’s second law of motion that net force is equal to an object’s mass times its acceleration as well as the kinematic equations when the acceleration of the object of interest is constant help us understand object motion.

Knowing about this topic largely boils down to understanding these three ideas. One: in an isolated system, energy is conserved. Two: Newton’s second law helps us understand object dynamics. And three: the kinematic equations can help us understand object motion further. Let’s get some practice using these ideas in a few examples.

A bullet has a mass of 2.60 grams and moves horizontally at a speed of 335 meters per second as it collides with a stack of eight pine boards, each 0.750 inches thick. The bullet decelerates as it penetrates the boards. And the bullet comes to rest just as it has moved the full distance through all eight boards. Find the average force exerted by the boards on the bullet. Assume that the motion of the boards due to the bullet’s impact is negligible.

Looking to solve here for the average force exerted by the boards on a bullet, we can call that force 𝐹. And we’ll start on our solution by drawing a diagram of this situation. We have in this scenario a bullet with an initial speed 𝑣 sub 𝑖 of 335 meters per second encountering a stack of eight pine boards. Each of the boards has a thickness 𝑇 of 0.750 inches. And we’re told that by the time the bullet makes it all the way through the last board, it comes to a stop.

Knowing all this, we want to solve for the average force 𝐹 that the boards exert on the bullet as it decelerates. Seeking to solve for average force may remind us of Newton’s second law of motion, which says that the net force on an object equals its mass times its acceleration 𝑎. Using this law, we want to solve for 𝐹. And we’re given 𝑚 in the problem statement. But we don’t yet know the acceleration of the bullet as it comes to a stop. However, if we assume that acceleration 𝑎 is constant, then that means the kinematic equations apply for describing the motion of the decelerating bullet.

Looking over these four equations of motion, we see that the second one helps us solve for what we want to know — acceleration — in terms of values we’re given. Written in terms of the variables for our particular situation, we can say that the final speed of the bullet squared is equal to its initial speed squared plus two times its acceleration times the distance it travels eight times the thickness of a board.

We know the bullet ends up at rest. So 𝑣 sub 𝑓 is equal to zero. So when we rearrange to solve for 𝑎, it equals negative 𝑣 sub 𝑖 squared all over two times eight 𝑇. We can substitute this expression for acceleration in for 𝑎 in our equation for force. And we understand that even though the minus sign implies that the bullet is decelerating, which is true. Since we want to solve for the average force the boards exert on the bullet, which will be positive, we’ll change that to a plus sign.

Looking at this expression, we were given the mass of the bullet 𝑚 in the problem statement as well as the initial speed of the bullet 𝑣 sub 𝑖. We’re also told the thickness 𝑇 of the boards. But that thickness is currently expressed in inches and we like to convert it to meters. 0.750 inches is approximately 1.905 centimeters. When we plug our values into this expression, we’re careful to use a mass in units of kilograms and a distance of the thickness of each board in units of meters. Entering this expression on our calculator, it comes out to 960 newtons. To two significant figures, that’s the average force that the boards exert on the bullet.

Now, let’s look at an example involving work and energy.

A shot-putter accelerates 7.27 kilograms shot from rest to 14.0 meters per second in 1.20 seconds, lifting it 0.800 meters as he does so. Calculate the power output in watts of the shot-putter as he does this, excluding the power produced to accelerate his body. Calculate the power output in metric horsepower of the shot-putter as he does this, excluding the power produced to accelerate his body. In metric horsepower, one horsepower equals 735.5 watts.

In this two-part problem, first, we wanna calculate the power output in units of watts and second the power output in units of horsepower. We can label those values 𝑃 sub 𝑤 and 𝑃 sub ℎ𝑝, respectively. Let’s start by drawing a diagram.

The shot-putter lifts the shot of mass 𝑚 equals 7.27 kilograms up through a vertical distance of ℎ equals 0.800 meters over a time we’ve called Δ𝑡 of 1.20 seconds. Over this time, the shot which started at rest acquires a speed — we’ve called 𝑣 sub 𝑖 — of 14.0 meters per second. The question is how much power did the shot-putter have to exert to move the shot like this.

Considering power, we recall that power 𝑝 is equal to work done over a time interval 𝑡. So we can write that the power output by the shot-putter in watts is equal to the total work done on the shot divided by Δ𝑡. We can recall that the work-energy theorem tells us that the change in an object’s kinetic energy ΔKE is equal to the work done on that object. But we know that in this case, that’s not the whole story when it comes to the work done by the shot-putter. And the reason is that in addition to the speed it’s acquired, the shot is raised at vertical distance ℎ through this process.

Combining this fact with the work-energy theorem, we can write the total work the shot-putter does on the shot is equal to the change in kinetic energy of the shot plus its mass times the acceleration due to gravity times the vertical height through which it was raised. We’ll treat the acceleration due to gravity as exactly 9.8 meters per second squared. When we further recall that the kinetic energy of an object is one-half its mass times its speed squared, we can rewrite our expression for the work done on the shot as one-half the shot’s mass times its speed 𝑣 sub 𝑖 squared plus 𝑚𝑔ℎ.

So we see the work done on the shot is equal to its mechanical energy: its kinetic plus its potential energy. Factoring out the shot’s mass, we now have an expression for work that we can substitute into our equation for 𝑃 sub 𝑤. 𝑃 sub 𝑤 equals 𝑚 over Δ𝑡 times the quantity 𝑣 sub 𝑖 squared over two plus 𝑔 times ℎ. We know all these values or they’re given as constants. So we’re ready to plug in and solve for 𝑃 sub 𝑤. When we plug in for the values of 𝑚, Δ𝑡, 𝑣 sub 𝑖, 𝑔, and ℎ and enter this expression on our calculator, we find that to three significant figures it’s 641 watts. That’s the power output by the shot-putter on the shot.

In part two, we want to solve for the same amount of power, but now expressed in units of metric horsepower. To calculate 𝑃 sub ℎ𝑝, we’ll take 𝑃 sub 𝑤 and divide it by the conversion factor between watts and metric horsepower. When we substitute in the value for 𝑃 sub 𝑤, we see the units of watts cancel out. We’re left with units of horsepower. And the value to three significant figures is 0.872 horsepower. So the shot-putter used almost nine tenths the power of a horse in accelerating the shot.

Let’s summarize what we’ve learnt about mechanical energy and dynamics. We’ve seen that mechanical energy or ME for short equals the sum of an object’s potential and kinetic energy. Written as an equation, we can write that ME is equal to KE plus PE. We’ve also seen that mechanical energy is conserved in an isolated system; that is, a system where energy is neither added nor taken away. And finally, we’ve seen that dynamics involves object motion caused by forces. And this motion may be described using Newton’s second law of motion as well as the kinematic equations of motion.

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