In physics, work is the energy transferred to or from an object via the application of force along a displacement. In its simplest form, it is often represented as the product of force and displacement. A force is said to do positive work if (when applied) it has a component in the direction of the displacement of the point of application. A force does negative work if it has a component opposite to the direction of the displacement at the point of application of the force.
For example, when a ball is held above the ground and then dropped, the work done by the gravitational force on the ball as it falls is equal to the weight of the ball (a force) multiplied by the distance to the ground (a displacement). When the force F is constant and the angle between the force and the displacement s is θ, then the work done is given by:
$W=Fs\cos {\theta }$
Work is a scalar quantity,^{[1]} so it has only magnitude and no direction. Work transfers energy from one place to another, or one form to another. The SI unit of work is the joule (J).
According to Jammer,^{[2]} the term work was introduced in 1826 by the French mathematician Gaspard-Gustave Coriolis^{[3]} as "weight lifted through a height", which is based on the use of early steam engines to lift buckets of water out of flooded ore mines. According to Rene Dugas, French engineer and historian, it is to Solomon of Caux "that we owe the term work in the sense that it is used in mechanics now".^{[4]}
Units
The SI unit of work is the joule (J), named after the 19th-century English physicist James Prescott Joule, which is defined as the work required to exert a force of one newton through a displacement of one metre.
The dimensionally equivalent newton-metre (N⋅m) is sometimes used as the measuring unit for work, but this can be confused with the measurement unit of torque. Usage of N⋅m is discouraged by the SI authority, since it can lead to confusion as to whether the quantity expressed in newton metres is a torque measurement, or a measurement of work.^{[5]}
The work W done by a constant force of magnitude F on a point that moves a displacement s in a straight line in the direction of the force is the product
$W=Fs$.
For example, if a force of 10 newtons (F = 10 N) acts along a point that travels 2 metres (s = 2 m), then W = Fs = (10 N) (2 m) = 20 J. This is approximately the work done lifting a 1 kg object from ground level to over a person's head against the force of gravity.
The work is doubled either by lifting twice the weight the same distance or by lifting the same weight twice the distance.
Work is closely related to energy. The work-energy principle states that an increase in the kinetic energy of a rigid body is caused by an equal amount of positive work done on the body by the resultant force acting on that body. Conversely, a decrease in kinetic energy is caused by an equal amount of negative work done by the resultant force. Thus, if the net work is positive, then the particle’s kinetic energy increases by the amount of the work. If the net work done is negative, then the particle’s kinetic energy decreases by the amount of the work.^{[6]}
From Newton's second law, it can be shown that work on a free (no fields), rigid (no internal degrees of freedom) body, is equal to the change in kinetic energy KE corresponding to the linear velocity and angular velocity of that body,
$W=\Delta KE.$
The work of forces generated by a potential function is known as potential energy and the forces are said to be conservative. Therefore, work on an object that is merely displaced in a conservative force field, without change in velocity or rotation, is equal to minus the change of potential energy PE of the object,
$W=-\Delta PE.$
These formulas show that work is the energy associated with the action of a force, so work subsequently possesses the physical dimensions, and units, of energy.
The work/energy principles discussed here are identical to electric work/energy principles.
Constraint forces
Constraint forces determine the object's displacement in the system, limiting it within a range. For example, in the case of a slope plus gravity, the object is stuck to the slope and, when attached to a taut string, it cannot move in an outwards direction to make the string any 'tauter'. It eliminates all displacements in that direction, that is, the velocity in the direction of the constraint is limited to 0, so that the constraint forces do not perform work on the system.
For a mechanical system,^{[7]} constraint forces eliminate movement in directions that characterize the constraint. Thus the virtual work done by the forces of constraint is zero, a result which is only true if friction forces are excluded.^{[8]}
Fixed, frictionless constraint forces do not perform work on the system,^{[9]} as the angle between the motion and the constraint forces is always 90°.^{[9]} Examples of workless constraints are: rigid interconnections between particles, sliding motion on a frictionless surface, and rolling contact without slipping.^{[10]}
Work is a scalar quantity,^{[1]} so it has only magnitude and no direction. Work transfers energy from one place to another, or one form to another. The SI unit of work is the joule (J).
According to Jammer,^{}[2] the term work was introduced in 1826 by the French mathematician Gaspard-Gustave Coriolis^{[3]} as "weight lifted through a height", which is based on the use of early steam engines to lift buckets of water out of flooded ore mines. According to Rene Dugas, French engineer and historian, it is to Solomon of Caux "that we owe the term work in the sense that it is used in mechanics now".^{[4]}
Units
The SI unit of work is the joule (J), named after the 19th-century English physicist James Prescott Joule, which is defined as the work required to exert a force of one newton through a displacement of one metre.
The dimensionally equivalent newton-metre (N⋅m) is sometimes used as the measuring unit for work, but this can be confused with the measurement unit of torque. Usage of N⋅m is discouraged by the SI authority, since it can lead to confusion as to whether the quantity expressed in newton metres is a torque measurement, or a measurement of work.^{[5]}
The work W done by a constant force of magnitude F on a point that moves a displacement s in a straight line in the direction of the force is the product
$W=Fs$.
For example, if a force of 10 newtons (F = 10 N) acts along a point that travels 2 metres (s = 2 m), then W = Fs = (10 N) (2 m) = 20 J. This is approximately the work done lifting a 1 kg object from ground level to over a person's head against the force of gravity.
The work is doubled either by lifting twice the weight the same distance or by lifting the same weight twice the distance.
Work is closely related to energy. The work-energy principle states that an increase in the kinetic energy of a rigid body is caused by an equal amount of positive work done on the body by the resultant force acting on that body. Conversely, a decrease in kinetic energy is caused by an equal amount of negative work done by the resultant force. Thus, if the net work is positive, then the particle’s kinetic energy increases by the amount of the work. If the net work done is negative, then the particle’s kinetic energy decreases by the amount of the work.^{[6]}
From The SI unit of work is the joule (J), named after the 19th-century English physicist James Prescott Joule, which is defined as the work required to exert a force of one newton through a displacement of one metre.
The dimensionally equivalent newton-metre (N⋅m) is sometimes used as the measuring unit for work, but this can be confused with the measurement unit of torque. Usage of N⋅m is discouraged by th
The dimensionally equivalent newton-metre (N⋅m) is sometimes used as the measuring unit for work, but this can be confused with the measurement unit of torque. Usage of N⋅m is discouraged by the SI authority, since it can lead to confusion as to whether the quantity expressed in newton metres is a torque measurement, or a measurement of work.^{[5]}
The work W done by a constant force of magnitude F on a point that moves a displacement s in a straight line in the direction of the force is the product
$W$
For example, if a force of 10 newtons (F = 10 N) acts along a point that travels 2 metres (s = 2 m), then W = Fs = (10 N) (2 m) = 20 J. This is approximately the work done lifting a 1 kg object from ground level to over a person's head against the force of gravity.
The work is doubled either by lifting twice the weight the same distance or by lifting the same weight twice the distance.
Work is closely related to energy. The work-energy principle states that an increase in the kinetic energy of a rigid body is caused by an equal amount of positive work done on the body by the resultant force acting on that body. Conversely, a decrease in kinetic energy is caused by an equal amount of negative work done by the resultant force. Thus, if the net work is positive, then the particle’s kinetic energy increases by the amount of the work. If the net work done is negative, then the particle’s kinetic energy decreases by the amount of the work.^{The work is doubled either by lifting twice the weight the same distance or by lifting the same weight twice the distance.
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Work is closely related to energy. The work-energy principle states that an increase in the kinetic energy of a rigid body is caused by an equal amount of positive work done on the body by the resultant force acting on that body. Conversely, a decrease in kinetic energy is caused by an equal amount of negative work done by the resultant force. Thus, if the net work is positive, then the particle’s kinetic energy increases by the amount of the work. If the net work done is negative, then the particle’s kinetic energy decreases by the amount of the work.^{[6]}
From Newton's second law, it can be shown that work on a free (no fields), rigid (no internal degrees of freedom) body, is equal to the change in kinetic energy KE corresponding to the linear velocity and angular velocity of that body,
The work of forces generated by a potential function is known as potential energy and the forces are said to be conservative. Therefore, work on an object that is merely displaced in a conservative force field, without change in velocity or rotation, is equal to minus the change of potential energy PE of the object,
$W=-\Delta PE.$physical dimensions, and units, of energy.
The work/energy principles discussed here are identical to electric work/energy principles.
Constraint forces
Constraint forces determine the object's displacement in the system, limiting it within a range. For example, in the case of a slope plus gravity, the object is stuck to the slope and, when attached to a taut string, it cannot move in an outwards direction to make the string any 'tauter'. It eliminates all displacements in that direction, that is, the velocity in the direction of the constraint is limited to 0, so that the constraint forces do not perform work on the system.
For a mechanical system,^{[7]} constraint forces eliminate movement in directions that characterize the constraint. Thus the mechanical system,^{[7]} constraint forces eliminate movement in directions that characterize the constraint. Thus the virtual work done by the forces of constraint is zero, a result which is only true if friction forces are excluded.^{[8]}
Fixed, frictionless constraint forces do not perform work on the system,^{[9]} as the angle between the motion and the constraint forces is always 90°.^{[9]} Examples of workless constraints are: rigid interconnections between particles, sliding motion on a frictionless surface, and rolling contact without slipping.^{[10]}
For example, in a pulley system like the Atwood machine, the internal forces on the rope and at the supporting pulley do no work on the system. Therefore work need only be computed for the gravitational forces acting on the bodies. Another example is the centripetal force exerted inwards by a string on a ball in uniform circular motionsideways constrains the ball to circular motion restricting its movement away from the centre of the circle. This force does zero work because it is perpendicular to the velocity of the ball.
The magnetic force on a charged particle is F = qv × B, where q is the charge, v is the velocity of the particle, and B is the magnetic field. The result of a cross product is always perpendicular to both of the original vectors, so F ⊥ v. The dot product of two perpendicular vectors is always zero, so the work W = F ⋅ v = 0, and the magnetic force does not do work. It can change the direction of motion but never change the speed.
For moving objects, the quantity of work/time (power) is integrated along the trajectory of the point of application of the force. Thus, at any instant, the rate of the work done by a force (measured in joules/second, or watts) is the scalar product of the force (a vector), and the velocity vector of the point of application. This scalar product of force and velocity is known as instantaneous power. Just as velocities may be integrated over time to obtain a total distance, by the fundamental theorem of calculus, the total work along a path is similarly the time-integral of instantaneous power applied along the trajectory of the point of application.^{[11]}
Work is the result of a force on a point that follows a curve X, with a velocity v, at each instant. The small amount of work δW that occurs over an instant of time dt is calculated as
$$X, with a velocity v, at each instant. The small amount of work δW that occurs over an instant of time dt is calculated as
where the F ⋅ v is the power over the instant dt. The sum of these small amounts of work over the trajectory of the point yields the work,
where C is the trajectory from x(t_{1}) to x(t_{2}). This integral is computed along the trajectory of the particle, and is therefore said to be path dependent.
If the force is always directed along this line, and the magnitude of the force is F, then this integral simplifies to
$W=\int _{C}F\,ds$
where s is displacement along the line. If F is constant, in addition to being directed along the line, then the integral simplifies further to
$W=\int _{C}F\,ds=F\int _{C}ds=Fs$
where s is the displacement of the point along the line.
This calculation can be generalized for a constant force that is not directed along the line, followed by the particle. In this case the dot productF ⋅ ds = F cos θds, where θ is the angle between the force vector and the direction of movement,^{[11]} that is
$If\; the\; force\; is\; always\; directed\; along\; this\; line,\; and\; the\; magnitude\; of\; the\; force\; is$F, then this integral simplifies to
where s is displacement along the line. If F is constant, in addition to being directed along the line, then the integral simplifies further to
$W=\int _{C}F\,ds=F\int _{C}ds=Fs$
where s is the displacement of the point along the line.
This calculation can be generalized for a constant force that is not directed along the line, followed by the particle. In this case the dot productF ⋅ ds = F cos θds, where θ is the angle between the force vector and the direction of movement,^{[11]} that is
This calculation can be generalized for a constant force that is not directed along the line, followed by the particle. In this case the dot productF ⋅ ds = F cos θds, where θ is the angle between the force vector and the direction of movement,^{[11]} that is
When a force component is perpendicular to the displacement of the object (such as when a body moves in a circular path under a central force), no work is done, since the cosine of 90° is zero.^{[6]} Thus, no work can be performed by gravity on a planet with a circular orbit (this is ideal, as all orbits are slightly elliptical). Also, no work is done on a body moving circularly at a constant speed while constrained by mechanical force, such as moving at constant speed in a frictionless ideal centrifuge.
Work done by a variable force
Calculating the work as "force times straight path segment" would only apply in the most simple of circumstances, as noted above. If force is changing, or if the body is moving along a curved path, possibly rotating and not necessarily rigid, then only the path of the application point of the force is relevant for the work done, and only the component of the force parallel to the application point velocity is doing work (positive work when in the same direction, and negative when in the opposite direction of the velocity). This component of force can be described by the scalar quantity called scalar tangential component (F cos(θ), where θ is the angle between the force and the velocity). And then the most general definition of work can be formulated as follows:
Work of a force is the line integral of its scalar tangential component along the path of its application point.
If the force varies (e.g. co
Calculating the work as "force times straight path segment" would only apply in the most simple of circumstances, as noted above. If force is changing, or if the body is moving along a curved path, possibly rotating and not necessarily rigid, then only the path of the application point of the force is relevant for the work done, and only the component of the force parallel to the application point velocity is doing work (positive work when in the same direction, and negative when in the opposite direction of the velocity). This component of force can be described by the scalar quantity called scalar tangential component (F cos(θ), where θ is the angle between the force and the velocity). And then the most general definition of work can be formulated as follows:
Work of a force is the line integral of its scalar tangential component along the path of its application point.
If the force varies (e.g. compressing a spring) we need to use calculus to find the work done. If the force is given by F(x) (a function of force couple results from equal and opposite forces, acting on two different points of a rigid body. The sum (resultant) of these forces may cancel, but their effect on the body is the couple or torque T. The work of the torque is calculated as
$dW=\mathbf {T} \cdot {\vec {\omega }}dt,$
where the T ⋅ ω is the power over the instant δt. The sum of these small amounts of work over the trajectory of the rigid body yields the work,
$W={\int}_{{t}_{1}}^{{t}_{2}}$
where the T ⋅ ω is the power over the instant δt. The sum of these small amounts of work over the trajectory of the rigid body yields the work,