Work (physics)

Does anyone know the amount of linear force that has to be applied to a baseball to accelerate it from 0 to 100 mph?

Now assuming that the force will be applied by a baseball pitcher, in the direction of motion, for a length of 2.5 feet (30 inches) from the beginning of the acceleration phase to the end release, what would the average force that is applied to the ball by the pitcher's hand have to be?

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In physics, mechanical work is the amount of energy transferred by a force. Like energy, it is a scalar quantity, with SI units of joules. Heat conduction is not considered to be a form of work, since there is no macroscopically measurable force, only microscopic forces occurring in atomic collisions. In the 1830s, the French mathematician Gaspard-Gustave Coriolis coined the term work for the product of force and distance.

Positive and negative signs of work indicate whether the object exerting the force is transferring energy to some other object, or receiving it. A baseball pitcher, for example, does positive work on the ball, but the catcher does negative work on it. Work can be zero even when there is a force. The centripetal force in uniform circular motion, for example, does zero work because the kinetic energy of the moving object doesn't change. Likewise, when a book sits on a table, the table does no work on the book, because no energy is transferred into or out of the book.

When the force is constant and along the same line as the motion, the work can be calculated by multiplying the force by the distance, $W=Fd$ (letting both F and d have positive or negative signs, according to the coordinate system chosen). When the force does not lie along the same line as the motion, this can be generalized to the scalar product of force and displacement vectors.

The baseball pitcher does work on the ball by transferring energy into it.

Calculation

In the simplest case, that of a body moving in a steady direction, and acted on by a constant force parallel to that direction, the work is given by the formula

$W=FD\,\!$             (1)

where

F is the portion of the force acting in the same direction as the motion, and

D is the distance traveled by the object. Note that distance is a scalar quantity and so, too, is work.

The work is taken to be negative when the force opposes the motion. More generally, the force and distance are taken to be vector quantities, and combined using the dot product:

$W={\vec {F}}\cdot {\vec {D}}=|{\vec {F}}||{\vec {D}}|\cos \phi$             (2)

where $\phi \,$ is the angle between the force and the displacement vector. This formula holds true even when the object changes its direction of travel throughout the motion.

In situations in which the force changes over time, and/or the direction of motion changes over time, equation (1) is not directly applicable. However, under mild restrictions, it is possible to divide the motion into small steps, such that the force and motion are well approximated as being constant for each step, and then to express the overall work as the sum over these steps. This is formalized by the following line integral, which can be taken as a rather general definition of work:

$W_{C}:=\int _{C}{\vec {F}}\cdot \mathrm {d} {\vec {s}}$             (3)

where:

C is the path or curve traversed by the object;

${\vec {F}}$ is the force vector;

${\vec {s}}$ is the position vector.

It must be emphasized that $W_{C}\,$ is explicitly a function of the path $C\!$. If work were a potential it would depend only on the endpoints of the path, but this is not the case; in general the work $W_{C}\,$ depends on every detail of the path $C\!$. As a related matter, it is not proper to write $\mathrm {d} W={\vec {F}}\cdot \mathrm {d} {\vec {s}}$ nor $\mathrm {d} W=\!$ anything (except perhaps in trivial cases, which we exclude from further consideration). That's because d(anything) is guaranteed to be an exact differential, whereas the correct expression ${\vec {F}}\cdot \mathrm {d} {\vec {s}}$ is an inexact differential. It is fairly common to see "$\mathrm {d} W\,$" used as shorthand for ${\vec {F}}\cdot \mathrm {d} {\vec {s}}$, but this must be considered highly informal and mathematically unjustifiable. Certainly there is no function "$W\,$" that can be differentiated to give ${\vec {F}}\cdot \mathrm {d} {\vec {s}}$.

Equation (3) readily explains how a nonzero force can do zero work. The simplest case is where the force is always perpendicular to the direction of motion, making the integrand always zero (viz. circular motion). However, even if the integrand sometimes takes nonzero values, it can still integrate to zero if it is sometimes negative and sometimes positive.

The possibility of a nonzero force doing zero work exemplifies the difference between work and a related quantity: impulse (the integral of force over time). Impulse measures change in a body's momentum, a vector quantity sensitive to direction, whereas work considers only the magnitude of the velocity. For instance, as an object in uniform circular motion traverses half of a revolution, its centripetal force does no work, but it transfers a nonzero impulse.

Units

Main article: work (thermodynamics)

The SI derived unit of work is the joule (J), which is defined as the work done by a force of one newton acting over a distance of one meter. This definition is based on Sadi Carnot's 1824 definition of work as "weight lifted through a height", which is based on the fact that early steam engines were principally used to lift buckets of water, though a gravitational height, out of flooded ore mines. The dimensionally equivalent newton-meter (N·m) is sometimes used instead; however, it is also sometimes reserved for torque to distinguish its units from work or energy.

Non-SI units of work include the erg, the foot-pound, the foot-poundal, and the liter-atmosphere.

Types of work

Forms of work that are not evidently mechanical in fact represent special cases of this principle. For instance, in the case of "electrical work", an electric field does work on charged particles as they move through a medium.

One mechanism of heat conduction is collisions between fast-moving atoms in a warm body with slow-moving atoms in a cold body. Although colliding atoms do work on each other, the force averages to nearly zero in bulk, so conduction is not considered to be mechanical work.

PV work

Chemical thermodynamics studies PV work, which occurs when the volume of a fluid changes. PV work is represented by the following equation:

$W_{C}=-\int _{C}P\,\mathrm {d} V$           (4)

where:

• W = work done on the system
• P = external pressure
• V = volume

Like all work functions, PV work is dependent on the path $C\!$. (The path in question is a curve in the Euclidean space specified by the fluid's pressure and volume, and infinitely many such curves are possible.) From a thermodynamic perspective, this fact implies that PV work is not a state function. This means that the differential $P\,\mathrm {d} V$ is an inexact differential.

To say the same thing more mathematically, $P\,\mathrm {d} V$ is not an exact one-form. There is no function (0-form) $W$ which is the potential of $P\,\mathrm {d} V$ and if $W\,$ doesn't exist then $\mathrm {d} W\!$ doesn't exist either. If it did exist, we would be able to use Stokes Theorem to find the value of the $W\!$ function at the boundary of the path (that is, the initial and final points) and therefore the work would be a state function. This impossibility is consistent with the fact that it does not make sense to refer to the work on a point; work presupposes a path.

PV work is often measured in the (non-SI) units of liter-atmospheres, where 1 L·atm = 101.3 J.

Mechanical energy

Main article: Mechanical energy

The mechanical energy of a body is that part of its total energy which is subject to change by mechanical work. It includes kinetic energy and potential energy. Some notable forms of energy that it does not include are thermal energy (which can be increased by frictional work, but not easily decreased) and rest energy (which is constant so long as the rest mass remains the same).

The relation between work and kinetic energy

If an external work W acts upon a body, causing its kinetic energy to change from E_k1 to E_k2, then:

$W=\Delta E_{k}=E_{k2}-E_{k1}\,\!$

Also, if we substitute the equation for kinetic energy that states $E_{k}=(1/2)mv^{2}$, we then get:

$W=\Delta ((1/2)mv^{2})=(1/2)mv_{2}^{2}-(1/2)mv_{1}^{2}$

Conservation of mechanical energy

The principle of conservation of mechanical energy states that, if a system is subject only to conservative forces (e.g. only to a gravitational force), its mechanical energy remains constant.

For instance, if an object with constant mass is in free fall, the total energy of position 1 will equal that of position 2.

$(E_{k}+E_{p})_{1}=(E_{k}+E_{p})_{2}\,\!$

where

• $E_{k}$ is the kinetic energy, and
• $E_{p}$ is the potential energy.

The external work will usually be done by the friction force between the system on the motion or the internal-non conservative force in the system or loss of energy due to heat.

External links

• Work - a chapter from an online textbook
• Work, Power, Kinetic Energy on Project PHYSNET