Work, Energy & Power
Work, Energy & Power
In this module we are going to cover the topics of work, energy and power. These concepts are very closely related, so it is important to understand the relationships and the differences between each of these variables. Work, energy and power also have direct connections with exercise physiology, several of theses connections will be discussed in this module. It is important to understand that there are corresponding physiological terms, so make sure that you distinguish between mechanical work energy and power (which we will cover here) and metabolic work, energy and power (which we will touch on, but you will learn in considerably more detail in exercise physiology).
Work
We will start this module by discussing the mechanical variable of work. Work is the product of force and the amount of displacement in the direction of that force. Work is also the means by which energy is transferred from one object to another. Given that there are these two definitions of work, there are two corresponding formulas for calculating work:
U = Fd
Where U = mechanical work (J), F = force (N) and d = displacement in the direction of the force (m)
U = ∆ energy = energyfinal - energyinitial
Where U = mechanical work (J), ∆ (delta) = change in, and energy refers to any type of mechanical energy (in this course we will only use potential energy (PE) and kinetic energy (KE) (J)
The unit of work is the joule, which is a newton (the unit of force) multiplied by a meter (the unit of distance).
The above image shows a weightlifter performing mechanical work. Specifically, mechanical work is done in lifting the barbell. If the weight lifter were taller, he would have to produce proportionally more work to press the barbell over his head (the d in the U=Fd equation would be greater).
Sample Problems
1. How much work does it take to lift a 30 kg suitcase onto the table, 1 meter high?
In this case, the force in the U = Fd equation is Fweight.
U = Fweightd
U = (30 kg * 9.81 m/s2)(1 m)
U = 294.3 J
2. How much worked is required to push a crate 10 m across a floor with a force of 250 N?
U = Fd
U = (250 N)(10 m)
U = 2500 J
3. How much work does gravity do on a 1 kg book that it has pulled off a 2-meter shelf?
In this case, the force in the U = Fd equation is Fweight.
U = Fweightd
U = (10 kg * 9.81 m/s2)(2 m)
U = 19.62 J
Working at an Advantage
Often we are limited by the amount of force we can apply. For example, even if you are putting your “full weight” into a wrench you are limited by your mass and gravity. Fortunately, we have a variety of tools, such as ramps, levers, pulleys, etc. that allow us to manipulate the variables of the U = Fd equation, such that the same amount of work is produced, but by applying a smaller force over a larger distance
Example: Working at an Advantage—Ramps
Let’s look at a ramp as an example that helps us work at an advantage. Ramps allow us to exert a smaller force over a larger distance to achieve the same change in gravitational potential energy (height raised).
Example:
If friction is neglected an identical amount of work is performed when lifting the box straight up or moving it up a ramp.
Lifting the box straight up:
U = Fd
U = (10 kg * 9.81 m/s2) * 5 m
U = 490.5 J
Using the ramp:
F = sin 30o *10 kg *9.81 m/s2
F = 49.05N
sin30o = o/h
sin30o = (5 m)/d
d = 10 m
U = Fd
U = 49.05 N * 10 m
U = 490.5 J
If we consider friction:
Fn= cos30o *10 kg *9.81m/s2
Fn = 84.96 N
Ffriction = μ Fn
Ffriction = 0.4 * 84.96 N = 33.98 N
From the previous example: Fdown = 49.05 N
So the total force required to move the box up the ramp must be greater than 33.98 N + 49.05 N = 83.03 N
This means that the total amount of work increases:
U = Fd
U = 83.03 N*10 m
U = 830.3 J
However, often a ramp can still seem easier. This is because we are often limited in the amount of force that can be applied. Additionally, the example here utilizes a very high coefficient of friction. Using a low friction ramp or rollers/wheels, would significantly reduce the amount of work required to overcome friction.
Work & Energy Expenditure
Mechanical work can be used to estimate energy expenditure. In more simple terms, performing mechanical work makes you tired. In fact, calories are another unit of work, 1 Joule is 0.239 calories, or 0.000239 kCal (food equivalent). As touched on previously, there are physiological variables that correspond with to the mechanical variables that we are covering in this module. For example, let’s look at the comparison of mechanical work and metabolic work:
- Metabolic work—metabolic energy consumption
- Mechanical work:
- External work—energy required to move the COM
- Internal work—energy required to move the limbs relative to the COM
A primary method to describe the relationship between mechanical work and metabolic work is the variable of efficiency, which is the ratio of mechanical work to metabolic work:
Efficiency = mechanical work/metabolic work
The efficiency of human muscle is ~26%. Knowing this value we can make calculations of mechanical work and use this information to estimate metabolic work. Let's take for example our example of moving the box up the ramp:
How many Calories are expended to push the box up the slope?
U = 830.3 J
U = 830.3 J * 0.000239 Calories/J
U = 0.198 kCal
In the course of a work day a worker moves 500 boxes up the slope. How many kCals are expended?
U = 0.198 kCal * 500 boxes
U = 99 kCal
This seems really low, why? This method only accounts for the movement of the 5 kg box, it does not take into consideration the movement of the worker, returning to the start of the ramp, etc.
Energy
The next variable that we will talk about is energy, which is the capacity to do work. Energy comes in many forms: heat, light, sound, chemical, etc., but in biomechanics we are primarily concerned with mechanical energy. Mechanical energy comes in two forms: potential and kinetic energy, the units of of which are kg*m2/s2 = Nm = J. Note that work and energy share the same unit, which we will cover in more detail in the section on the work-energy relationship.
Law of Conservation of Energy
Before we dive into the details of energy, it is important to cover the Law of Conservation of Energy, which states that the total energy of an isolated system cannot change. Specifically, the total energy (in all forms) in a “closed” system remains constant. However, while the total energy remains constant in a closed system, energy can be transformed from one form to another. Take a roller coaster for example, while this isn’t a perfectly closed system, it does illustrate the conservation of energy. At the top of the track the coaster is high, so it has a great amount of potential energy, the coaster is moving slowly at this point indicating that it has low kinetic energy. As the coaster rolls down the track it loses height, therefore its’ potential energy decreases, but it gains velocity indicating that that kinetic energy is increasing. This indicates that the potential energy is being transferred to kinetic energy. At the bottom of the track the potential energy is lowest because the coaster is at its’ lowest, yet kinetic energy is maximized because the velocity of the car is greatest. Moving up the next curve the coaster’s kinetic energy will be transformed to potential resulting in the car slowing down (decreasing kinetic energy) as the kinetic energy is transferred to potential energy to give the car height (increase potential energy).
Energy Loss
It is important to note that the law of conservation of energy applies only to closed systems, but in reality, rarely is a system every closed. In most systems there are avenues for energy to be lost. One of the most common sources of energy loss is to heat, which is a viable (and common) path for energy flow. Heat is the product of friction, as well as many chemical, electrical processes. The downside of heat is that it is hard to make heat energy do anything for you because it is too disordered to extract useful work. Alternatively, we can produce a considerable amount of work from other energy sources, for example, kinetic energy of hammer can drive nail and the potential energy in compressed spring can produce motion.
Mechanical Energy
The two main types of mechanical energy that we will discuss in this course are potential energy and kinetic energy. We’ll start by discussing Potential Energy, which is the energy is an object’s capacity to do work based on position. There are two main types of potential energy:
- Gravitational Potential Energy—potential energy due to an object’s position relative to earth.
- Elastic (sprain) Energy—energy due to deformation of an object.
In this course we will focus primarily on the topic of gravitational potential energy, which is often simply referred to as potential energy, but we will touch on the topic of elastic energy.
Gravitational Potential Energy
Gravitational potential energy is energy due to an object’s position relative to earth. The equation for gravitational potential energy is:
PE = mgh
Where PE = potential energy (J), m = mass (kg), g = acceleration due to gravity = 9.81 m/s2, h = vertical height (m)
It is important to note that gravitational potential energy only depends on the vertical height above the earth/ground, it does not depend on how that height was reached. For example, in the image below the potential energy of the 10 N ball is the same (30 J) in all three cases because the work done in elevating it the 3 m height is the same whether it is (a) lifted straight up with 10 N of force, (b) pushed with 6 N of force up a 5 m incline, or (c) lifted with 10 N up each 1 m stair.
Sample problem:
Roger Federer’s tennis ball has a mass of 0.3 kg. If he holds the ball above the ground at a height of 2.0 m to serve, what is its gravitational potential energy (GPE)?
PE = mgh = 0.3 kg x 9.81 m/s2 x 2.0 m = 5.89 J
Elastic Energy
Elastic energy is the energy due to the deformation of an object. The calculation of elastic energy is beyond the scope of this course, but we will estimate elastic energy storage when we cover gait (running in particular) at the end of the course. However, let’s look at some examples of elastic energy.
A pole vaulter stores strain energy in the pole when loading the pole by planting it in the box. This energy is what propels the vaulter up and over the bar (work is performed). Additionally, a very important feature of tendons is that they store and return energy elastically. We’ll see in the gait section of the course that this is very important for reducing the metabolic cost of running. This is also the basis of spring-like running prosthetics. These prosthetics are compressed, which stores energy elastically and is used to propel the runner forward.
Kinetic Energy
The next type of energy that we will discuss is kinetic energy, which is the energy an object has due to its’ motion. Hence, a moving object has the capacity to do work because of its’ motion. The equation for kinetic energy is as follows:
KE = ½ mv2
Where KE = kinetic energy (J), m = mass (kg) and v = velocity (m/s)
Sample problems:
1. What is the Kinetic Energy of a 90 kg sprinter running at a speed of 15 m/s?
KE = ½ mv2
KE = ½ (90 kg) (15m/s)2
KE = 10,125 J
2. Missy Diwater, the former platform diver for the Ringling Brother's Circus, had a kinetic energy of 12,000 J just prior to hitting the bucket of water. If Missy's mass is 40 kg, then what is her speed?
KE = 0.5 m v2
12 000 J = (0.5) (40 kg) v2
300 J = (0.5) v2
600 J = v2
v = 24.5 m/s
Work Energy Relationship
Work and energy are closely related, which you might have guessed from the fact that they share a common unit (Joules) and each is indicated in the definition of the other. Specifically, work is the means by which energy is transferred from one object or system (or the change in energy) and energy is the capacity to do work. In human movement we are often concerned with changing the velocity of the body (or a body part), which would mean changing kinetic energy, or changing the height of the body (or a body part), which would mean changing potential energy.
Let’s also put some numbers to this to illustrate the work-energy relationship.
A weight lifter bench presses a bar with a weight of 50 kg over a distance of 0.7 m. How much work was done?
In this case we will use the equation:
U = Fd
In this case the force is the weight of the bar, which is calculated as:
Fw = mg
Fw = 50 kg * 9.81 m/s2 = 490.5 N
Now we multiply this force by the height (distance moved) to calculate the work done:
U = Fd
U = 490.5 N * 0.7 m = 343.35 J
Alternatively we can calculate the change in energy:
U = Δenergy = energyfinal – energyinitial
First, we need to decide what type of mechanical energy we are going to use (KE or PE)? When deciding this we need to think about what is changing in terms of the end state of the object. Are we changing the height of the object? In which case we would be examining PE. Or are we changing the velocity of the object? In which case we would be examining KE. In this case, while velocity does change during the bench press, the height of the barbell changes from start to end. Therefore, we can plug in PE as the energy in our equation:
U = PEf – PEi
We aren’t given a starting height for the barbell, so we can set the initial height to 0, and the final height to 0.7 m, which gives us:
U = (50 kg * 9.81 m/s2 * 0.7 m) - (50 kg * 9.81 m/s2 * 0 m)
U = (50 kg * 9.81 m/s2 * 0.7 m) - 0 = 343.45 J
Here you can see that we obtained the exact same result for the amount of work performed on the barbell regardless of if we calculated the work done as force times displacement or the change in energy. However, it is important to note that you may not always have enough information to calculate work in both ways, so you may be limited to one equation or the other.
Power
The final variable that we will discuss in this module is power, which is the rate of doing work. The unit of power is the Watt, which is a Joule per second. Power can be calculated in several ways, first simply as work divided by time:
P = U/t
Where P = power (W), U = work (J) and t = time (s)
Since we also know that work can be calculated as the product of force and displacement, we can replace U by Fd:
P = Fd/t
Where P = power (W), F = Force (N), d = displacement in the direction of the force (m), and t = time (s)
We also know that displacement divided by time is velocity, so we can replace d/t with v:
P = Fv
Where P = power (W), F = Force (N) and v = velocity (m/s)
Sample problem:
- When doing a chin-up, a student lifts her 62.0 kg body a distance of 0.25 meters in 2 seconds. What is the power delivered by the student's biceps?
P = U/t
U = Fd
U = 62 kg * 9.81 m/s2 * 0.25 m
U = 152.1 J
P = U/t
P = 152.1 J / 2 s
P = 76.03 W
Power is an important component of many human movements, because often it is not only the amount of force that is applied, but also the rate of force application. Take the power clean for example, the lifter cannot generate enough force to simply lift the weight from the ground to their shoulders, but by producing a large force and then quickly moving under the bar (high velocity), they are able to perform the lift. This combination of force and velocity highlights the importance of power in many dynamic movements and is the rationale behind plyometric training, which aims to improve both the force and velocity (and hence power) components of human movement.
Let’s also specifically look at muscular power. We covered the force velocity relationship of human muscle previously and we saw that for concentric muscle actions as force increased velocity decreased and vice versa. Hence, if we plot muscle power as a function of velocity, we see that peak power is obtained at intermediates of force and velocity. If the force is too high, velocity will be low and subsequently so will the power output of the muscle. Similarly, if the velocity is too high, force is too low, and we also see low muscle power. Therefore, to maximize muscle power we need intermediate values of force and velocity.
Efficiency
Lastly let’s look at the connection between mechanical work/power and metabolic energy consumption. In particular, we are going to look at the variable of efficiency, which is the ratio of mechanical power (or work) to metabolic power (or work):
Efficiency = Pmech /Pmet
or
Efficiency = Umech /Umet
Where the efficiency of human muscle is 26%, Pmech = mechanical power (W), Pmet = metabolic power (W), Umech = mechanical work (J) and Umet = metabolic work (J).
This is a convenient relationship because, as we have seen, mechanical work can be fairly easily calculated and we also have a value “known” value of the efficiency of human muscle, which is ~26%. If we know both efficiency and mechanical power/work we can use this to estimate metabolic power and metabolic power can be used to estimate the rate of oxygen consumption (VO2), given that:
Pmet = VO2 * (20.1 J/ml O2)
Where Pmet = metabolic power (W) and VO2 = the volume of oxygen consumption (ml O2/s). It is important to note that in exercise physiology the volume of oxygen consumption is often reported as the amount of oxygen consumed (ml O2) per kilogram of body weight per minute (ml O2/kg/min).
Let’s look at an example of how mechanical power can be used to estimate oxygen consumption. A 70 kg cyclist is riding their 5 kg bike up a 10o slope at 5 m/s. We’ll start by calculating their mechanical power output:
We aren’t given a displacement, so of our three equations for calculating mechanical power, we will need to use:
Pmech = Fv
In this case, the force that they have to overcome is the downward component of their weight force (Fdown).
Fdown = sin10o * (70 kg + 5 kg)* 9.81 m/s2
Fdown = 127.8 N
Plugging this into the equation Pmech = Fv, gives us:
Pmech = Fv
Pmech = 127.8 N * 5 m/s
Pmech = 639 W
Then solving for metabolic power using the efficiency equation:
Efficiency = Pmech/Pmet
26% = Pmech/Pmet
Pmet = 639 W /0.26 = 2,458 W
Then using metabolic power to estimate oxygen consumption:
Pmet = VO2 * (20.1 J/ml O2)
2,458W = VO2 * (20.1 J/ml O2)
VO2 = 122.3 mlO2 /s
Finally, converting the oxygen consumption into the units that are normally used by exercise physiologists (mlO2/kg/min):
VO2 = 122.3 ml O2 /s *60 s/min /75 kg = 98 ml O2/kg/min
If you are familiar with VO2max values of top endurance athletes you will recognize that this is really high, so let’s look at why this is the case. First, it is important to realize that this calculation assumes that the work being performed is all produced aerobically. In this case the cyclist is riding at 11.2 mph (5 m/s) up a 20% grade (10o slope), so it is highly likely that at least part of the mechanical work being performed is coming from anaerobic energy sources, which are ignored in this calculation.
References
When no attribution is noted, images are from the Public Domain or the work of the author.