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Work Energy and Power

Energy, work, and power are three interconnected concepts in physics. However, did you know that we use these concepts daily? Energy, a necessary component of life, is found in everything we eat. Eating provides our bodies with the energy needed to regulate internal functions such as repairing cells or body tissue, building muscle, and maintaining homeostasis. We may not know it but energy is always present as it allows our bodies to do “work” and provides the “power” needed for us to function properly. Therefore, let us use this example as a starting point in understanding work, energy, and power, and introduce definitions and examples that help expand our knowledge on the topic.

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Jetzt kostenlos anmeldenEnergy, work, and power are three interconnected concepts in physics. However, did you know that we use these concepts daily? Energy, a necessary component of life, is found in everything we eat. Eating provides our bodies with the energy needed to regulate internal functions such as repairing cells or body tissue, building muscle, and maintaining homeostasis. We may not know it but energy is always present as it allows our bodies to do “work” and provides the “power” needed for us to function properly. Therefore, let us use this example as a starting point in understanding work, energy, and power, and introduce definitions and examples that help expand our knowledge on the topic.

Work, energy, and power are terms heard in everyday life. However, their technical definitions may not be as widely familiar to us, so let us define each term, starting with energy.

**Energy** is a system’s ability to do work.

Now from this definition, we are led straight to *work, *no pun intended.

**Work** is the amount of energy transferred due to an object moving some distance because of an external force.

Energy and work both have the same corresponding SI unit, joules denoted by \( \mathrm{J} \). On the other hand, power is a quantity that encompasses energy and work.

**Power** is the rate at which energy (work done) is transferred or transformed with respect to time.

Power has a SI unit of watts, denoted by \( \mathrm{W} \). Note that all three quantities are scalar, meaning they have magnitude but no direction.

Now as we have defined these terms, we must discuss their corresponding formulas. This can be complicated as each term has multiple formulas, however, do not worry. We will discuss one term at a time and define each of its corresponding formulas. Let's start with energy. Energy exists in many forms, but all energy is classified as kinetic or potential.

**Kinetic energy** is the energy associated with motion.

An easy way to remember this definition is to remember that the word *kinetic* means motion. Now the corresponding formula to this definition is

$$K=\frac{1}{2}mv^2$$

where \( m \) is mass measured in \( \mathrm{kg} \) and \( v \) is velocity measured in \( \mathrm{\frac{m}{s}}. \)

This formula represents translational kinetic energy, which is energy due to linear motion.

By contrast, potential energy focuses on position rather than motion.

**Potential Energy** is energy due to an object's position.

The mathematical formula for potential energy varies depending on circumstances within a system. Therefore, let's go through some different forms and discuss their formulas. One of the most common forms taught early on when studying physics is gravitational potential energy.

**Gravitational potential energy** is the energy of an object due to its vertical height.

Gravitational potential energy corresponds to the formula $$U=mgh,$$

where \( m \) is mass measured in \( \mathrm{kg} \), \( g \) is the acceleration due to gravity, and \( h \) is height measured in \( \mathrm{m} \).

Note that mass and height are directly related to gravitational potential energy. The larger the mass and height values are, the larger the potential energy value will be.

Now gravitational potential energy can also be defined in terms of calculus. The **calculus definition** describes the relationship between conservative forces exerted on a system and gravitational potential energy, **\( \Delta U =-\int \vec{F}(x)\cdot \mathrm{d}x. \)** This integral is equal to the work required to move between two points and describes the change in gravitational potential energy. If we use in conjunction with our knowledge that gravitational potential energy is equal to \( U=mgh \), we can show how the calculus definition is used to derive the simplest equation for gravitational potential energy.

$$\Delta U =-\int_{h_0}^h (-mg)\mathrm{d}h= (mgh-mgh_0)$$

If \( h_0 \) is set to zero to represent the ground, the equation becomes

$$\Delta U= mgh,$$

the simplest formula for determining gravitational potential energy.

It is important to note that the negative sign of the integral indicates that the force acting on the system is the negative derivative, \( F= -\frac{\mathrm{d}U(x)}{\mathrm{d}x} \), of the gravitational potential energy function, \( \Delta U \). This essentially means that it is the negative slope of a potential energy curve.

Another common form taught early on is elastic potential energy.

**Elastic potential energy **is the energy stored within an object due to its ability to be stretched or compressed.

Its corresponding mathematical formula is $$U=\frac{1}{2}k\Delta{x}^2,$$

where \( k \) is the spring constant and \( x \) is the compression or elongation of the spring. Elastic potential energy is directly related to the amount of stretch in a spring. The more stretch there is, the greater the elastic potential energy is.

**Two Spherical Masses**

Now, the final formula for potential energy we will discuss is for a system of two spherical masses, such as planets, stars, and the moon. The potential energy formula corresponding to this type of system is

$$U=-\frac{Gm_1m_2}{r},$$

where \( G \) is the gravitational constant and \( m \) is mass. The negative sign here indicates the bound state of a mass once it approaches a large body. This means that the mass is stuck until enough energy is provided for it to become unbound.

As we have defined various types of energy, we also must discuss a key concept corresponding to energy. This concept is the **conservation of energy **which states that energy cannot be created nor destroyed.

**Conservation of energy: **The total mechanical energy, which is the sum of all potential and kinetic energy, of a system remains constant when excluding dissipative forces.

Dissipative forces are nonconservative forces, such as friction or drag forces, in which work is dependent on the path an object travels.

When calculating the total mechanical energy of a system, the following formula is used:

$$K_\mathrm{i} + U_\mathrm{i}= K_\mathrm{f} + U_\mathrm{f}$$

where \( K \) is kinetic energy and \( U \) is potential energy. This equation does not apply to a system consisting of a single object because, in that particular type of system, objects only have kinetic energy. This formula is only used for systems in which interactions between objects are caused by *conservative forces*, forces in which work is independent of the path an object travels because the system may then have both kinetic and potential energy. Now if we have an isolated system, the total energy of the system remains constant because nonconservative forces are excluded and the net work done on the system is equal to zero. However, if a system is open, energy is transformed. Although the amount of energy in a system remains constant, energy will be converted into different forms when work is done. Work done on a system causes changes in the total mechanical energy due to internal energy.

**Total internal energy** is the sum of all energies comprising an object.

Total internal energy changes due to dissipative forces. These forces cause the internal energy of a system to increase while causing the total mechanical energy of the system to decrease. For example, a box, undergoing a frictional force, slides along a table but eventually comes to stop because its kinetic energy transforms into internal energy. Therefore, to calculate the total mechanical energy of a system in which work is done, the formula

\( K_\mathrm{i} + U_\mathrm{i}= K_\mathrm{f} + U_\mathrm{f} + {\Delta{E}} \), must be used to account for this transfer of energy. Note that \( {\Delta{E}} \) represents the work done on the system which causes a change in internal energy.

As energy leads us to work, let's now define and discuss the mathematical formulas corresponding to work. Formulas for work vary depending on circumstances within a system. As a result, we will define work in terms of three forces, a constant force, a conservative force, and a variable force.

If a system consists of a constant force along a linear path, the corresponding work formula is

$$W=Fd,$$

where \( F \) is force measured in \( \mathrm{N} \) and \( d \) is distance measured in \( \mathrm{m}. \)

If the work done on a system is independent of the path due to a conservative force, the corresponding work formula is

$$W_{\mathrm{conservative}}={-\Delta U} = {\Delta K},$$

where \( -\Delta U \) is the negative change in potential energy and \( \Delta{K} \) is the change in kinetic energy.

A conservative force is a force in which work is only dependent on the final displacement of an object.

When work is done due to a variable force, we define it in terms of calculus where its corresponding formula is

$$W=\int \vec{F}\cdot \mathrm{d}\vec{r}.$$

The integral, in this formula, takes two important factors into account:

- Force may vary in magnitude and direction,
- The path may also vary in direction.

From this formula, we should also recognize that a dot product is another term for vector product indicating that both quantities are vector quantities. As a result, the dot-product results in a scalar quantity of magnitude \( W= Fr\cos \theta \).

Power, the term that encompasses both energy and work, comes in different forms. As a result, we will discuss instantaneous power, average power, and peak power. Let's start with instantaneous power.

**Instantaneous power** is the power measured at a specific moment in time.

The corresponding formula to this definition is $$P_\mathrm{inst}=\frac{\mathrm{d}W}{\mathrm{d}t}=Fv,$$

where \( F \) is force and \( v \) is velocity. Instantaneous power also can be expressed by the formula

$$P_\mathrm{inst}=Fv \cos\theta,$$

which is derived from the equation for average power. Using the equation for average power and rewriting work in terms of its definition, we get the following:

\begin{align}P_\mathrm{avg} &=\frac{W}{\Delta t} \\P_\mathrm{avg} &=\frac{Fx\cos\theta}{\Delta t} \\P_\mathrm{inst} &= Fv\cos\theta\end{align}

Note that \( v \) is the result of \( \frac{x}{t} \).

In contrast, average power considers longer amounts of time.

**Average Power** is the power measured within a long time period.

The corresponding mathematical formula is \( P_\mathrm{avg}=\frac{W}{\Delta t}=\frac{\Delta E}{\Delta t}. \) Lastly, **peak power** is defined as the maximum instantaneous power of a system within a long time period.

Work, energy, and power are interconnected concepts in physics. Work and energy are related by the work-energy theorem. This theorem, derived from Newton's second law, states that the net work done on an object is equal to its change in kinetic energy. Power and work are related because power is the rate at which work is done. Energy is also related to power because power is the rate at which energy is used or transferred. However, although they are interconnected concepts, it is important to understand that they are three separate terms with individual definitions and formulas.

Efficiency is defined as a comparison ratio between the input and useful output quantities of a system. This ratio can be expressed in terms of work, energy, and power. To calculate the efficiency of work, use the formula $$\mathrm{Efficiency} =\frac{W_\mathrm{out}}{W_\mathrm{in}} \times 100%. $$ For power efficiency, use $$\mathrm{Efficiency} =\frac{P_\mathrm{out}}{P_\mathrm{in}} \times 100%,$$ and for energy efficiency, use $$\mathrm{Efficiency} =\frac{E_\mathrm{out}}{E_\mathrm{in}} \times 100%.$$ Note that efficiency is a scalar quantity and has no units.

To solve work, energy, and power problems, one can use their corresponding equations and apply them to different problems. As we have defined work, energy, and power and discussed their relation to one another, let us work through some examples to gain familiarity with the equations. Note that before solving a problem, we must always remember these simple steps:

- Read the problem and identify all variables given within the problem.
- Determine what the problem is asking and what formulas are needed.
- Apply the necessary formulas and solve the problem.
- Draw a picture if necessary to provide a visual aid.

Let's apply our new knowledge of work, energy, and power to the following three examples.

A sled is pulled across the snow with a horizontal force of \( 63.0\,\mathrm{N} \) for \( 28.0\,\mathrm{m} \) at a constant velocity. What is the work done on the sled? If the sled is then pulled with a force of \( 60.0\,\mathrm{N} \) at an angle of \( 45.0 \) degrees for \( 20.0\,\mathrm{m} \), calculate the work done.

Based on the problem, we are given the following:

- force
- displacement
- the angle,\( \theta \)

As a result, we can identify and use the equation, \( W=Fd \), to solve the first part of this problem. Therefore, our calculations are:

\begin{align}W &= Fd \\W &= (63\,\mathrm{N})(28\,\mathrm{m}) \\W &= 1764\,\mathrm{J}\end{align}

The equation, \( W= Frcos \theta \), will be used for the last part of the problem. As a result, our calculations are:

\begin{align}W &= Fr\cos \theta \\W &= (60\,\mathrm{N})(20\,\mathrm{m}) \cos (45^{\circ}) \\W &=849\,\mathrm{J}\end{align}

As we have calculated work, let's solve for power in the example below.

If a force of \( 350.0\,\mathrm{N} \) is required to keep an object moving with a constant speed of \( 5.3\,\mathrm{\frac{m}{s}} \), how much power is needed to maintain this motion?

Based on the problem, we are given the following:

- force
- velocity

As a result, we can identify and use the equation, \( P_\mathrm{inst}=Fv \), to solve this problem. Therefore, our calculations are:

\begin{align}P_\mathrm{inst} &= Fv \\P_\mathrm{inst} &= (350\,\mathrm{N})\left(5.3\,\mathrm{\frac{m}{s}}\right) \\P_\mathrm{inst} &= 1855\,\mathrm{W}\end{align}

Finally, let's use our knowledge of gravitational potential energy to solve for velocity.

A \( 7.50\,\mathrm{kg} \) object is dropped from a height of \( 8.00\,\mathrm{m} \) above the ground. What is the velocity of the object as it strikes the ground?

Based on the problem, we are given the following:

- mass
- height

As a result, we can identify and use the equation, \( K_\mathrm{i} + U_\mathrm{i}= K_\mathrm{f} + U_\mathrm{f} \), to solve this problem. Therefore, our calculations are:

\begin{align}K_\mathrm{i} + U_\mathrm{i} &= K_\mathrm{f} + U_\mathrm{f} \\0 + U_\mathrm{i} &= K_\mathrm{f} + 0 \\mgh &= \frac{1}{2}mv^2 \\(7.5\,\mathrm{kg})\left(9.81\,\mathrm{\frac{m}{s^2}}\right)(8\,\mathrm{m}) &= \frac{1}{2}(7.5\,\mathrm{kg})v^2 \\v^2 &= 156.96\,\mathrm{\frac{m^2}{s^2}} \\v &= 12.5 \mathrm{\frac{m}{s}} \\\end{align}

Note that initial kinetic energy is equal to zero because the object was at rest and final kinetic energy equals zero because the velocity of the object will be zero after it strikes the ground.

- Energy is defined as a system’s ability to do work.
- Work is defined as the amount of energy transferred due to an object moving some distance because of an external force.
- Power is the rate at which energy is transferred or transformed with respect to time.
- Work, energy, and power are scalar quantities.
- Energy exists in many forms: but all energy is classified as either kinetic or potential energy.
- Kinetic energy is the energy associated with motion.
- Potential Energy is energy due to an object's position.
- The formula for potential energy is defined in terms of calculus, elastic potential, gravitational, and gravitational potential energy of two spherical masses.
- Conservation of energy states that the sum of all potential and kinetic energy, of a system, remains constant when excluding dissipative forces.
- The formula for work is defined in terms of calculus, constant forces, and conservative forces.
- The formula for power is defined in terms of instantaneous power and average power.
- Work, energy, and power are interconnected concepts with individual definitions and formulas.
- Efficiency is defined as a comparison ratio between the input and useful output quantities of a system.

- Figure 1: Human Body (https://www.pexels.com/photo/photo-of-woman-studying-anatomy-3059750/) by RF_Studio (https://www.pexels.com/@rethaferguson/) is licensed by CC0 1.0 Universal (CC0 1.0).
- Figure 2: Pulling a Sled- StudySmarter Originals
- Figure 3: Dropping an object- StudySmarter Originals

What kind of force gives a system potential energy?

A conservative force.

What is the conservative force that gives a skydiver in free fall potential energy?

Gravity.

What is the equation for the change in potential energy?

\[\Delta U=-\int_{x_1}^{x_2}F(x)\,\mathrm{d}x.\]

How do you find the force function if you are given the function for potential energy?

You find the force function by taking minus the derivative of the function for potential energy: \(F(x)=-\frac{\mathrm{d} U(x)}{\mathrm{d}x}\).

Which of the following is a conservative force?

Gravity.

Which of the following is the correct definition of potential energy?

Potential energy is energy that comes from the position and internal configuration of two or more objects in a system.

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