Physical Applications

Work

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.

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

$\delta W=\mathbf {F} \cdot d\mathbf {s} =\mathbf {F} \cdot \mathbf {v} dt$

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,

$W=\int _{t_{1}}^{t_{2}}\mathbf {F} \cdot \mathbf {v} dt=\int _{t_{1}}^{t_{2}}\mathbf {F} \cdot {\tfrac {d\mathbf {s} }{dt}}dt=\int _{C}\mathbf {F} \cdot d\mathbf {s} ,$

where C is the trajectory from x(t₁) to x(t₂). 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 product F ⋅ ds = F cos θ ds, where θ is the angle between the force vector and the direction of movement, that is

$W=\int _{C}\mathbf {F} \cdot d\mathbf {s} =Fs\cos \theta .$

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. 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. compressing a spring) we need to use calculus to find the work done. If the force is given by F(x) (a function of x) then the work done by the force along the x-axis from a to b is:

$W=\int _{a}^{b}\mathbf {F(s)} \cdot d\mathbf {s}$

Work and potential energy

The scalar product of a force F and the velocity v of its point of application defines the power input to a system at an instant of time. Integration of this power over the trajectory of the point of application, C = x(t), defines the work input to the system by the force.

Path dependence

Therefore, the work done by a force F on an object that travels along a curve C is given by the line integral:

$W=\int _{C}\mathbf {F} \cdot d\mathbf {x} =\int _{t_{1}}^{t_{2}}\mathbf {F} \cdot \mathbf {v} dt,$

where dx(t) defines the trajectory C and v is the velocity along this trajectory. In general this integral requires the path along which the velocity is defined, so the evaluation of work is said to be path dependent.

The time derivative of the integral for work yields the instantaneous power,

${\frac {dW}{dt}}=P(t)=\mathbf {F} \cdot \mathbf {v} .$

Path independence

If the work for an applied force is independent of the path, then the work done by the force, by the gradient theorem, defines a potential function which is evaluated at the start and end of the trajectory of the point of application. This means that there is a potential function U(x), that can be evaluated at the two points x(t₁) and x(t₂) to obtain the work over any trajectory between these two points. It is tradition to define this function with a negative sign so that positive work is a reduction in the potential, that is

$W=\int _{C}\mathbf {F} \cdot \mathrm {d} \mathbf {x} =\int _{\mathbf {x} (t_{1})}^{\mathbf {x} (t_{2})}\mathbf {F} \cdot \mathrm {d} \mathbf {x} =U(\mathbf {x} (t_{1}))-U(\mathbf {x} (t_{2})).$

The function U(x) is called the potential energy associated with the applied force. The force derived from such a potential function is said to be conservative. Examples of forces that have potential energies are gravity and spring forces.

In this case, the gradient of work yields

$\nabla W=-\nabla U=-\left({\frac {\partial U}{\partial x}},{\frac {\partial U}{\partial y}},{\frac {\partial U}{\partial z}}\right)=\mathbf {F} ,$

and the force F is said to be "derivable from a potential."

Because the potential U defines a force F at every point x in space, the set of forces is called a force field. The power applied to a body by a force field is obtained from the gradient of the work, or potential, in the direction of the velocity V of the body, that is

$P(t)=-\nabla U\cdot \mathbf {v} =\mathbf {F} \cdot \mathbf {v} .$

Work by gravity

Gravity F = mg does work W = mgh along any descending path

In the absence of other forces, gravity results in a constant downward acceleration of every freely moving object. Near Earth's surface the acceleration due to gravity is g = 9.8 m⋅s⁻² and the gravitational force on an object of mass m is F'_g = mg. It is convenient to imagine this gravitational force concentrated at the center of mass of the object.

If an object with weight mg is displaced upwards or downwards a vertical distance y₂ − y₁, the work W done on the object is:

$W=F_{g}(y_{2}-y_{1})=F_{g}\Delta y=mg\Delta y$

where F_g is weight (pounds in imperial units, and newtons in SI units), and Δy is the change in height y. Notice that the work done by gravity depends only on the vertical movement of the object. The presence of friction does not affect the work done on the object by its weight.

Work by gravity in space

The force of gravity exerted by a mass M on another mass m is given by

$\mathbf {F} =-{\frac {GMm}{r^{3}}}\mathbf {r} ,$

where r is the position vector from M to m.

Let the mass m move at the velocity v; then the work of gravity on this mass as it moves from position r(t₁) to r(t₂) is given by

$W=-\int _{\mathbf {r} (t_{1})}^{\mathbf {r} (t_{2})}{\frac {GMm}{r^{3}}}\mathbf {r} \cdot d\mathbf {r} =-\int _{t_{1}}^{t_{2}}{\frac {GMm}{r^{3}}}\mathbf {r} \cdot \mathbf {v} dt.$

Notice that the position and velocity of the mass m are given by

$\mathbf {r} =r\mathbf {e} _{r},\qquad \mathbf {v} ={\frac {d\mathbf {r} }{dt}}={\dot {r}}\mathbf {e} _{r}+r{\dot {\theta }}\mathbf {e} _{t},$

where e_r and e_t are the radial and tangential unit vectors directed relative to the vector from M to m, and we use the fact that $d\mathbf {e} _{r}/dt={\dot {\theta }}\mathbf {e} _{t}.$ Use this to simplify the formula for work of gravity to,

$W=-\int _{t_{1}}^{t_{2}}{\frac {GmM}{r^{3}}}(r\mathbf {e} _{r})\cdot ({\dot {r}}\mathbf {e} _{r}+r{\dot {\theta }}\mathbf {e} _{t})dt=-\int _{t_{1}}^{t_{2}}{\frac {GmM}{r^{3}}}r{\dot {r}}dt={\frac {GMm}{r(t_{2})}}-{\frac {GMm}{r(t_{1})}}.$

This calculation uses the fact that

${\frac {d}{dt}}r^{-1}=-r^{-2}{\dot {r}}=-{\frac {\dot {r}}{r^{2}}}.$

The function

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

is the gravitational potential function, also known as gravitational potential energy. The negative sign follows the convention that work is gained from a loss of potential energy.

Work by a spring

Forces in springs assembled in parallel

Consider a spring that exerts a horizontal force F = (−kx, 0, 0) that is proportional to its deflection in the x direction independent of how a body moves. The work of this spring on a body moving along the space with the curve X(t) = (x(t), y(t), z(t)), is calculated using its velocity, v = (v_x, v_y, v_z), to obtain

$W=\int _{0}^{t}\mathbf {F} \cdot \mathbf {v} dt=-\int _{0}^{t}kxv_{x}dt=-{\frac {1}{2}}kx^{2}.$

For convenience, consider contact with the spring occurs at t = 0, then the integral of the product of the distance x and the x-velocity, xv_xdt, over time t is (1/2)x². The work is the product of the distance times the spring force, which is also dependent on distance; hence the x² result.

Hooke's law for linear springs

Consider a simple helical spring that has one end attached to some fixed object, while the free end is being pulled by a force whose magnitude is F_s. Suppose that the spring has reached a state of equilibrium, where its length is not changing anymore. Let x be the amount by which the free end of the spring was displaced from its "relaxed" position (when it is not being stretched). Hooke's law states that

$F_{s}=kx$

or, equivalently,

$x={\frac {F_{s}}{k}}$

where k is a positive real number, characteristic of the spring. Moreover, the same formula holds when the spring is compressed, with F_s and x both negative in that case. According to this formula, the graph of the applied force F_s as a function of the displacement x will be a straight line passing through the origin, whose slope is k.

Hooke's law for a spring is sometimes, but rarely, stated under the convention that F_s is the restoring force exerted by the spring on whatever is pulling its free end. In that case, the equation becomes

$F_{s}=-kx$

since the direction of the restoring force is opposite to that of the displacement.

Work by a gas

$W=\int _{a}^{b}{P}dV$

Where P is pressure, V is volume, and a and b are initial and final volumes.

Hydrostatic pressure

In a fluid at rest, all frictional and inertial stresses vanish and the state of stress of the system is called hydrostatic. When this condition of {{{1}}} is applied to the Navier–Stokes equations, the gradient of pressure becomes a function of body forces only. For a barotropic fluid in a conservative force field like a gravitational force field, the pressure exerted by a fluid at equilibrium becomes a function of force exerted by gravity.

The hydrostatic pressure can be determined from a control volume analysis of an infinitesimally small cube of fluid. Since pressure is defined as the force exerted on a test area ($p={\frac {F}{A}}$, with p: pressure, F: force normal to area A, A: area), and the only force acting on any such small cube of fluid is the weight of the fluid column above it, hydrostatic pressure can be calculated according to the following formula:

$p(z)-p(z_{0})={\frac {1}{A}}\int _{z_{0}}^{z}dz'\iint _{A}dx'dy'\,\rho (z')g(z')=\int _{z_{0}}^{z}dz'\,\rho (z')g(z'),$

where:

• p is the hydrostatic pressure (Pa),
• ρ is the fluid density (kg/m³),
• g is gravitational acceleration (m/s²),
• A is the test area (m²),
• z is the height (parallel to the direction of gravity) of the test area (m),
• z₀ is the height of the zero reference point of the pressure (m).

For water and other liquids, this integral can be simplified significantly for many practical applications, based on the following two assumptions: Since many liquids can be considered incompressible, a reasonable good estimation can be made from assuming a constant density throughout the liquid. (The same assumption cannot be made within a gaseous environment.) Also, since the height h of the fluid column between z and z₀ is often reasonably small compared to the radius of the Earth, one can neglect the variation of g. Under these circumstances, the integral is simplified into the formula:

$p-p_{0}=\rho gh,$

where h is the height z − z₀ of the liquid column between the test volume and the zero reference point of the pressure. This formula is often called Stevin's law. Note that this reference point should lie at or below the surface of the liquid. Otherwise, one has to split the integral into two (or more) terms with the constant ρ_liquid and ρ(z′)_above. For example, the absolute pressure compared to vacuum is:

$p=\rho gH+p_{\mathrm {atm} },$

where H is the total height of the liquid column above the test area to the surface, and p_atm is the atmospheric pressure, i.e., the pressure calculated from the remaining integral over the air column from the liquid surface to infinity. This can easily be visualized using a pressure prism.

Hydrostatic force on submerged surfaces

The horizontal and vertical components of the hydrostatic force acting on a submerged surface are given by the following:

${\begin{aligned}F_{\mathrm {h} }&=p_{\mathrm {c} }A\\F_{\mathrm {v} }&=\rho gV\end{aligned}}$

where:

• p_c is the pressure at the centroid of the vertical projection of the submerged surface
• A is the area of the same vertical projection of the surface
• ρ is the density of the fluid
• g is the acceleration due to gravity
• V is the volume of fluid directly above the curved surface

Resources

• Work (physics), Wikipedia
• Hydrostatics, Wikipedia
• Hooke's Law, Wikipedia

Work Done by a Variable Force

• Work Done by a Variable Force by Krista King
• Work Done By a Variable Force Physics Problems by The Organic Chemistry Tutor
• Work Problems - Calculus by The Organic Chemistry Tutor

Work (Rope/Cable Problems)

• Ex 1: Integration Application - Work Lifting an Object by James Sousa, Math is Power 4U
• Ex 2: Integration Application - Work Lifting an Object and Cable by James Sousa, Math is Power 4U
• Ex: Find the Work Lifting a Leaking Bucket of Sand Given Mass by James Sousa, Math is Power 4U
• Ex: Find the Work Lifting a Leaking Bucket of Sand and Rope Given Mass by James Sousa, Math is Power 4U
• Finding Work using Calculus - The Cable/Rope Problem - Part b by patrickJMT
• Work done using a rope to lift a weight by patrickJMT
• Work Problems - Calculus by The Organic Chemistry Tutor

Work (Spring Problem)

• Ex: Find the Work Required to Stretch a Spring by James Sousa, Math is Power 4U
• Ex: Find the Force Required to Stretch a Spring by James Sousa, Math is Power 4U
• Work and Hooke's Law - Ex 1 by patrickJMT
• Work and Hooke's Law - Ex 2 by patrickJMT
• Work Done on Elastic Springs by Krista King
• Hooke's Law Physics, Basic Introduction, Restoring Force, Spring Constant by The Organic Chemistry Tutor
• Work Problems by The Organic Chemistry Tutor

Work (Pumping Fluid Out of a Tank)

• Ex: Determine the Work Required to Pump Water Out of a Circular Cylinder by James Sousa, Math is Power 4U
• Ex: Determine the Work Required to Pump Water Out of Trough (Isosceles Triangle) by James Sousa, Math is Power 4U
• Ex: Determine the Work Required to Pump Water Out of Trough (Quadratic Cross Section) by James Sousa, Math is Power 4U
• Calculating the Work Required to Drain a Tank by patrickJMT
• Work Problems - Calculus by The Organic Chemistry Tutor

Hydrostatic Pressure and Force

• Ex: Find the Hydrostatic Force on a Horizontal Plate (No Calculus) by James Sousa, Math is Power 4U
• Ex: Find the Hydrostatic Force on a Vertical Plane in the Shape of a Isosceles Triangle by James Sousa, Math is Power 4U
• Ex: Find the Hydrostatic Force on a Semicircle Window Submerged in Water by James Sousa, Math is Power 4U
• Ex: Find the Hydrostatic Force on a Dam in the Shape of a Degree 4 Polynomial by James Sousa, Math is Power 4U
• Hydrostatic Force - Basic Idea / Deriving the Formula by patrickJMT
• Hydrostatic Force - Complete Example #1 by patrickJMT
• Hydrostatic Force - Complete Example #2, Part 1 of 2 by patrickJMT
• Hydrostatic Force - Complete Example #2, Part 2 of 2 by patrickJMT
• Hydrostatic Pressure by Krista King
• Hydrostatic Force by Krista King
• Introduction to Pressure & Fluids by The Organic Chemistry Tutor
• Hydrostatic Force Problems by The Organic Chemistry Tutor

Licensing

Content obtained and/or adapted from:

• Work (physics), Wikipedia under a CC BY-SA license
• Hydrostatics, Wikipedia under a CC BY-SA license
• Hooke's Law, Wikipedia under a CC BY-SA license