# Work and Kinetic Energy

## Work done by a constant force

The term work, in physics, is a term whose meaning is different from everyday meaning. Consider an object being displaced along the x axis by a force, constant in magnitude and direction. We define the work W done on an object by constant force acting on the object as the product of the component of the force in the direction of the displacement and the magnitude of the displacement:

${\displaystyle W=F\times d=Fd\cos \theta }$

As example of the difference between the definition of work and the everyday understanding of the word, consider holding an heavy object for several minutes. Your arm at the end will be shacking for the amount of work done on the object. Instead, according to the definition given in physics, you would have done no work at all. This is because, you applied a force to support the object but you did not move it at all. Thus d=0 and the result of the product was equal to zero.

Note from definition that the work is null if the force is perpendicular to the displacement of the object as the ${\displaystyle \cos \theta =0}$ ${\displaystyle (\theta =90^{\circ })}$.

The sign of the work depends on the direction of F relative to d. In particular, the work is positive when the vector associated with the product F${\displaystyle \cos \theta }$ is in the same direction as the displacement. Contrary, it is negative when the direction of the vector associated with F${\displaystyle \cos \theta }$ is opposite the displacement. If the force F acts along the exact direction and verse of the displacement ${\displaystyle \cos \theta =1}$ and the equation gives

${\displaystyle W=Fd}$

Work is a scalar quantity, and its units are force multiplied by length. The SI unit of work is ${\displaystyle newton\cdot meter}$ (${\displaystyle N\cdot m}$). This combination of units has been given a name: the joule (J).

## Work done by a varying force

Consider now an object being displaced along the x axis by a varying force, , increasing x from ${\displaystyle x=x_{\text{i}}}$ to ${\displaystyle x=x_{\text{f}}}$. In such situation we can not use

${\displaystyle W=F\times d}$
to calculate the work done by the force F because the equation applies only when the force is constant in magnitude and direction. However, if we examine a very small fraction of the displacement when the x component of the force ${\displaystyle F_{\text{x}}}$ is approximately constant we can express the work done by the force as

${\displaystyle \Delta W=F_{\text{x}}\Delta x\qquad \Delta x=x_{\text{f}}-x_{\text{i}}}$

If we imagine the ${\displaystyle F_{\text{x}}}$ versus x curve divided in a large number of such intervals, then the total work along the displacement from ${\displaystyle x_{\text{i}}}$ to ${\displaystyle x_{\text{f}}}$ is approximately equal to the sum of such terms:

${\displaystyle W\thickapprox \sum _{x=x_{\text{i}}}^{x_{\text{f}}}F_{\text{x}}\Delta x}$

If the displacements approach zero, then the number of terms increases and the sum approaches a definite value equal to the area underneath the curve:

${\displaystyle \lim _{x\rightarrow 0}\sum _{x=x_{\text{i}}}^{x_{\text{f}}}F_{\text{x}}\Delta x=\int _{x_{\text{i}}}^{x_{\text{f}}}F_{\text{x}}dx\qquad \to \qquad W=\int _{x_{\text{i}}}^{x_{\text{f}}}F_{\text{x}}dx}$

This equation is reduced to

${\displaystyle W=F\times d}$
when the component ${\displaystyle Fd\cos \theta }$ is constant.

If more then one force is applied to an object the total work is the work done by the resultant force. Thus becomes

${\displaystyle \sum W=\int _{x_{\text{i}}}^{x_{\text{f}}}{\Big (}\sum F_{\text{x}}{\Big )}dx}$

## Work done by a varying force/Work done by a spring

A clear example of a variable force is the force that acts on an object connected to a spring that has been either stretched or compressed described by the following equation

${\displaystyle F_{\text{spring}}=-kx}$

where x represents the displacement of the object from its point of equilibrium (where the spring is unstretched/uncompressed), and k is the force constant of the spring. This force law for springs is called Hook's Law and it's valid only for small displacements. The k of a spring indicates its stiffness. To stiff spring are related big k values and, viceversa, to soft springs are related small k values.

The negative of the force indicates that the force produced from a spring is always directed the opposite way of the displacement. For this reason is called a restoring force. (For full details see Spring Force in newton mechanics [...]).

Suppose the object has been pushed to the left a distance ${\displaystyle x_{\text{max}}}$ from equilibrium. Let's calculate the work done by the spring force from the point ${\displaystyle x_{\text{i}}=-x_{\text{max}}}$ to the point of equilibrium ${\displaystyle x_{\text{f}}=0}$

${\displaystyle W_{\text{s}}=\int _{x_{\text{i}}}^{x_{\text{f}}}F_{\text{x}}dx=\int _{-x_{\text{max}}}^{0}(-kx)dx={\frac {1}{2}}kx_{\text{max}}^{2}}$

When we consider the work done by the spring force from ${\displaystyle x_{\text{i}}=0}$ to ${\displaystyle x_{\text{i}}=x_{\text{max}}}$ we find that ${\displaystyle W_{\text{s}}=-{\frac {1}{2}}k(x_{\text{max}})^{2}}$ because the spring force and the displacement have opposite direction. In general, if the object undergoes an arbitrary displacement form ${\displaystyle x=x_{\text{i}}}$ to ${\displaystyle x=x_{\text{i}}}$ the work done by the spring force is

${\displaystyle W_{\text{s}}=\int _{x_{\text{i}}}^{x_{\text{f}}}(-kx)dx={\frac {1}{2}}kx_{\text{i}}^{2}-{\frac {1}{2}}kx_{\text{f}}^{2}}$

Equation and describe the work done by the spring on the object. Consider now the work done on the spring by an external force. This force stretches the spring very slowly form ${\displaystyle x_{\text{i}}=0}$ to ${\displaystyle x_{\text{i}}=x_{\text{max}}}$. At any value of the displacement of the object the applied force ${\displaystyle {\textbf {F}}_{\text{app}}}$ is equal to the opposite of the spring force ${\displaystyle {\textbf {F}}_{\text{s}}}$, so that ${\displaystyle {\textbf {F}}_{\text{app}}=-(-kx)=kx}$. So, the work done by this applied force is

${\displaystyle W_{\text{F}}=\int _{0}^{x_{\text{f}}}F_{\text{app}}dx=\int _{0}^{x_{\text{f}}}kxdx={\frac {1}{2}}kx_{\text{f}}^{2}}$

This is the work equal to the negative of the work done by the spring force for his displacement.

## Kinetic energy ant the work-kinetic energy theorem

Solving motion problems involving complex forces can be difficult using Newton's second law. In fact, is easier to relate the speed of a moving object to its displacement under the influence of a force. If we can calculate the work done by that force its easy to extract the speed of the object.

Consider a constant net force ${\displaystyle \sum {\textbf {F}}}$ applied on an object of mass m. From Newton's second law the object is moving with constant acceleration a. If we call d the displacement of the object, the work done by the net force ${\displaystyle \sum {\textbf {F}}}$ is

${\displaystyle \sum {W}={\Big (}\sum {\textbf {F}}{\Big )}d=(ma)d}$

From the study of kinematic we know that if an object is moving with constant acceleration, then this relationships are valid

${\displaystyle d={\frac {1}{2}}(v_{\text{i}}+v_{\text{f}})t\qquad a={\frac {v_{\text{f}}-v_{\text{i}}}{t}}}$

Substituting these relationships in the equation we obtain a new relationship between the work done by a net force and the speed of the object on which the force is applied

${\displaystyle \sum {W}=m{\Big (}{\frac {v_{\text{f}}-v_{\text{i}}}{t}}{\Big )}{\frac {1}{2}}(v_{\text{i}}+v_{\text{f}})t}$

${\displaystyle \sum {W}={\frac {1}{2}}mv_{\text{f}}^{2}-{\frac {1}{2}}mv_{\text{i}}^{2}}$

The member ${\displaystyle {\frac {1}{2}}mv^{2}}$ represents the energy associated with the motion of the object, and it has been called kinetic energy.

In general the kinetic energy of a particle of mass m and speed v is

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

The kinetic energy is a scalar quantity and it has the same units as work, the joule. Is often useful to write the as

${\displaystyle \sum {W}=K_{\text{f}}-K_{\text{i}}=\Delta K}$

The is a very important result known as the work-kinetic energy theorem. It's important to note that when we apply this theorem we must include, in calculating the net work, all the forces applied to the particle. Also, from this theorem, we can deduce that the speed of the particle increases only if the work, done by the net force acting on it , is positive. This because the final kinetic energy is greater than the initial one. Contrary, the speed of the particle decreases only if the work is negative. In this configuration the final kinetic energy is smaller than the initial one, so the ${\displaystyle \Delta K}$ is negative.

The theorem expressed by the tell us two different characteristic of kinetic energy: the kinetic energy can describe the work that a particle can do in coming to rest, but also can be seen as the energy stored in a moving particle.

We derive the equation considering the work done only by a constant net force, but it also valid when the forces are variable. If we consider the x direction of ${\displaystyle \sum {F_{\text{x}}}}$, apply the Newton's secondo law, ${\displaystyle \sum {F_{\text{x}}}=ma_{\text{x}}}$, and use

${\displaystyle W=\int _{x_{\text{i}}}^{x_{\text{f}}}F_{\text{x}}dx}$
to express the work done by a varying force

${\displaystyle \sum W=\int _{x_{\text{i}}}^{x_{\text{f}}}{\Big (}\sum F_{\text{x}}{\Big )}dx=\int _{x_{\text{i}}}^{x_{\text{f}}}ma_{\text{x}}dx}$

If the resultant force is depending by x, the acceleration and speed also depend on x. Now consider the acceleration expressed in a not so usual way:

${\displaystyle a={\frac {dv}{dt}}={\frac {dv}{dx}}{\frac {dx}{dx}}=v{\frac {dv}{dx}}}$

Substituting this expression of a into the equation of ${\displaystyle \sum W}$ gives

${\displaystyle \sum W=\int _{x_{\text{i}}}^{x_{\text{f}}}mv{\frac {dv}{dx}}dx=\int _{v_{\text{i}}}^{v_{\text{f}}}mvdv}$

${\displaystyle \sum W={\frac {1}{2}}mv_{\text{f}}^{2}-{\frac {1}{2}}mv_{\text{i}}^{2}}$

## Situations involving kinetics friction

Adding the frictional forces into the analysis of the motion of an object sliding on an horizontal surface, we can describe the kinetic energy lost because of friction. Consider an object sliding on an horizontal rough surface we no forces applied to it. The only force acting on the object is the force of kinetic friction ${\displaystyle {\textbf {f}}_{\text{k}}}$ between the surface on which is sliding and the object itself, and it cause the object to accelerate in the negative x direction. Newton's second law gives ${\displaystyle -f_{\text{k}}=ma_{\text{x}}}$. Multiplying both members for d and using the equation ${\displaystyle v_{\text{xf}}^{2}-v_{\text{xi}}^{2}=2a_{\text{x}}d}$ for the motion of a particole under constant acceleration give ${\displaystyle -f_{\text{k}}d=(ma_{\text{x}})d={\frac {1}{2}}mv_{\text{xf}}^{2}-{\frac {1}{2}}mv_{\text{xi}}^{2}}$. In other terms:

${\displaystyle \Delta K=-f_{\text{k}}d}$

This equation tells that the difference in the kinetic energy of an object on which is applied a kinetic friction force is equal to ${\displaystyle -f_{\text{k}}d}$. Parts of this quantity goes into warming the surface of the object, part in warming the surface on which it slides.

When other forces, as well as the kinetic friction, concurrently act on an object the difference of work-kinetic energy theorem reads

${\displaystyle K_{\text{i}}+\sum W_{\text{other}}-f_{\text{k}}d=K_{\text{f}}}$

in other terms:

${\displaystyle \Delta K=\sum W_{\text{other}}-f_{\text{k}}d}$

where ${\displaystyle \sum W_{\text{other}}}$ is the work done by all the forces acting on the object except the kinetic friction one.

## Power

Power is the time rating of doing work. Consider an external force applied to a particle; the work done by this force in the interval ${\displaystyle \Delta t}$ is equal to W. The average power expended during this interval is:

${\displaystyle {\bar {P}}={\frac {W}{\Delta t}}}$

Istantaneous power can be defined as the limiting value of the average power as ${\displaystyle \Delta t}$ approaches zero:

${\displaystyle P=\lim _{\Delta t\to 0}{\frac {W}{\Delta t}}={\frac {dW}{dt}}}$

From the definition of the work done by a constant force

${\displaystyle W=F\times d}$
we have

${\displaystyle P={\frac {dW}{dt}}={\textbf {F}}\cdot {\textbf {v}}}$

where v=ds/dt

The SI unit of power is (J/s), also called watt (W) (after James Watt the inventor of the steam engine), where

${\displaystyle 1W=1J/s=1Kg\cdot m^{2}/s^{3}}$