# W=Integral Fdx – Work, Energy And Power

### Key Takeaways

Key PointsThe work done by a constant force of magnitude F on a point that moves a displacement d in the direction of the force is the product: W = Fd.Integration approach can be used both to calculate work done by a variable force and work done by a constant force.The SI unit of work is the joule; non- SI units of work include the erg, the foot-pound, the foot-poundal, the kilowatt hour, the litre-atmosphere, and the horsepower-hour.Key Termswork: A measure of energy expended in moving an object; most commonly, force times displacement. No work is done if the object does not move.

Đang xem: W=integral fdx

force: A physical quantity that denotes ability to push, pull, twist or accelerate a body, which is measured in a unit dimensioned in mass × distance/time² (ML/T²): SI: newton (N); CGS: dyne (dyn)

### Using Integration to Calculate the Work Done by Variable Forces

A force is said to do work when it acts on a body so that there is a displacement of the point of application in the direction of the force. Thus, a force does work when it results in movement.

The work done by a constant force of magnitude F on a point that moves a displacement Delta ext{x} in the direction of the force is simply the product

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ext{W}= ext{F}cdot Delta ext{x}

In the case of a variable force, integration is necessary to calculate the work done. For example, let’s consider work done by a spring. According to the Hooke’s law the restoring force (or spring force) of a perfectly elastic spring is proportional to its extension (or compression), but opposite to the direction of extension (or compression). So the spring force acting upon an object attached to a horizontal spring is given by:

mathbf{ ext{F}_{ ext{s}}}=- ext{k}mathbf{ ext{x}}

that is proportional to its displacement (extension or compression) in the x direction from the spring’s equilibrium position, but its direction is opposite to the x direction. For a variable force, one must add all the infinitesimally small contributions to the work done during infinitesimally small time intervals dt (or equivalently, in infinitely small length intervals dx=vxdt). In other words, an integral must be evaluated:

ext{W}_{ ext{s}}=int_0^ ext{t}mathbf{ ext{F}_{ ext{s}}}cdotmathbf{ ext{v}} ext{dt} =int_0^ ext{t} – ext{kx} hspace{3 pt} ext{v}_ ext{x} ext{dt} =int_{ ext{x}_ ext{o}}^ ext{x} – ext{kx} hspace{3 pt} ext{dx}= -frac{1}{2} ext{k}Delta ext{x}^2

This is the work done by a spring exerting a variable force on a mass moving from position xo to x (from time 0 to time t). The work done is positive if the applied force is in the same direction as the direction of motion; so the work done by the object on spring from time 0 to time t, is:

ext{W}_{ ext{a}}=int_0^ ext{t}mathbf{ ext{F}_{ ext{a}}}cdotmathbf{ ext{v}} ext{dt} =int_0^ ext{t}-mathbf{ ext{F}_{ ext{s}}}cdotmathbf{ ext{v}} ext{dt} = frac{1}{2} ext{k}Delta ext{x}^2

in this relation mathbf{ ext{F}_{ ext{a}}} is the force acted upon spring by the object. mathbf{ ext{F}_{ ext{a}}} and mathbf{ ext{F}_{ ext{s}}} are in fact action- reaction pairs; and mathbf{ ext{W}_{ ext{a}}} is equal to the elastic potential energy stored in spring.

### Using Integration to Calculate the Work Done by Constant Forces

The same integration approach can be also applied to the work done by a constant force. This suggests that integrating the product of force and distance is the general way of determining the work done by a force on a moving body.

Consider the situation of a gas sealed in a piston, the study of which is important in Thermodynamics. In this case, the Pressure (Pressure =Force/Area) is constant and can be taken out of the integral:

ext{W}=int_ ext{a}^ ext{b}{ ext{P}} ext{dV}= ext{P}int_ ext{a}^ ext{b} ext{dV}= ext{P} Delta ext{V}

Another example is the work done by gravity (a constant force) on a free-falling object (we assign the y-axis to vertical motion, in this case):

ext{W}=int_{ ext{t}_1}^{ ext{t}_2}mathbf{ ext{F}}cdotmathbf{ ext{v}} ext{dt} = int_{ ext{t}_1}^{ ext{t}_2} ext{mg} hspace{3 pt} ext{v}_ ext{y} ext{dt} = ext{mg} int_{ ext{y}_1}^{ ext{y}_2} ext{dy}= ext{mg}Delta ext{y}

Notice that the result is the same as we would have obtained by simply evaluating the product of force and distance.

### Units Used for Work

The SI unit of work is the joule (J), which is defined as the work done by a force of one newton moving an object through a distance of one meter.

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

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