Calculus

Applications of Integrals

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Force, Work and Energy

Work Done by a Force

Work in physics is defined as the product of the force and displacement.

When an object moves a distance \(\Delta x\) along a straight line as a result of action of a constant force \(F,\) the work done by the force is

\[W = F\Delta x.\]

This equation is called the constant force formula for work.

If the force is measured in Newtons \(\left( N \right)\) and distance is in meters \(\left( m \right),\) then work is measured in Joules \(\left( J \right).\)

If an object moves along a straight line from \(x = a\) to \(x = b\) under the influence of a variable force \(F\left( x \right),\) the work \(W\) done by the force is given by the definite integral

\[W = \int\limits_a^b {F\left( x \right)dx} .\]

Hooke's Law

Hooke's Law says that the force it takes to stretch or compress a spring \(x\) units from its natural (unstressed) length is

\[F = kx,\]

where \(F\) is the applied force, \(k\) is the spring constant, \(x\) is the displacement from the original length.

Hooke's Law
Figure 1.

Sometimes Hooke's Law is written as

\[F = - kx.\]

In this expression \(F\) no longer means the applied force, but rather means the equal and oppositely directed restoring force.

The work needed to stretch the spring from \(0\) to \(x\) is given by the integral

\[W = \int\limits_0^x {Fdx} = \int\limits_0^x {kxdx} = \frac{{k{x^2}}}{2}.\]

Newton's Law of Gravitation

According to Newton's law of universal gravitation, the gravitational force acting between two objects is given by

\[F = G\frac{{{m_1}{m_2}}}{{{x^2}}},\]

where \({m_1}\) and \({m_2}\) ate the masses of the objects, \(x\) is the distance between the centers of their masses, and \(G\) is the gravitational constant.

Coulomb's Law

Coulomb's law states that the force with which stationary electrically charged particles repel or attract each other is given by

\[F = k\frac{{{q_1}{q_2}}}{{{x^2}}},\]

where \({q_1}\) and \({q_2}\) are the signed magnitudes of the charges, \(x\) is the distance between the charges, and \(k\) is Coulomb's constant.

Kinetic Energy

The kinetic energy of a moving object is equal to

\[{E_k} = \frac{1}{2}m{v^2},\]

where \(m\) is the mass and \(v\) is the velocity of the object.

If a rigid body is rotating about any line through the center of mass with an angular velocity \(\omega,\) then it has rotational kinetic energy, which is given by the equation

\[{E_r} = \int {\frac{{{v^2}dm}}{2}} = \int {\frac{{{{\left( {r\omega } \right)}^2}dm}}{2}} = \frac{{{\omega ^2}}}{2}\int {{r^2}dm} = \frac{1}{2}I{\omega ^2},\]

where \(I\) is the moment of inertia and the integration is performed over all mass elements of the body.

Potential Energy

Potential energy is defined as the energy stored in an object. There are different kinds of potential energy, depending on the type of force. For example, the potential energy associated with gravitational force is called gravitational potential energy.

Assuming that the gravitational field near the Earth's surface is constant, the gravitational potential energy of a body is given by the formula

\[{E_p} = mgh,\]

where \(m\) is the mass, \(g\) is the acceleration of gravity, and \(h\) is the height.

Solved Problems

Click or tap a problem to see the solution.

Example 1

A force of \(50\,\text{N}\) is required to stretch a spring \(10\,\text{cm}\) beyond its natural length. How much work will be done stretching the spring \(20\,\text{cm}\) from its natural length?

Example 2

Calculate the work that is required to pump the water through an upper opening out of a vertical cylindrical barrel with base radius \(R\) and altitude \(H.\)

Example 3

Calculate the work that has to be done to raise a body of mass \(m\) from the Earth's surface to an altitude \(h.\) What is the work if the body is removed to infinity?

Example 4

A raindrop with initial mass \({m_0}\) starts falling from rest under the action of gravity and evenly evaporates losing every second mass equal to \(\mu.\) What is the work of gravity during the time from the beginning of the movement to the complete evaporation of the drop. Neglect air resistance.

Example 1.

A force of \(50\,\text{N}\) is required to stretch a spring \(10\,\text{cm}\) beyond its natural length. How much work will be done stretching the spring \(20\,\text{cm}\) from its natural length?

Solution.

According to Hooke's law,

\[F = kx.\]

Hence, the spring constant is equal to

\[k = \frac{F}{x} = \frac{{50}}{{0.1}} = 500\,\left( {\frac{\text{N}}{\text{m}}} \right),\]

where the displacement \(x\) is measured in meters: \({x = 10\,\text{cm} = 0.1\,\text{m}}\)

To find the work done by the external force, we integrate from \(x = 0\) to \(x = 20\,\text{cm} = 0.2\,\text{m}:\)

\[W = \int\limits_0^{0.2} {kxdx} = \int\limits_0^{0.2} {500xdx} = \left. {\frac{{500{x^2}}}{2}} \right|_0^{0.2} = \frac{{500 \times {{0.2}^2}}}{2} = 10\,\left( J \right).\]

Example 2.

Calculate the work that is required to pump the water through an upper opening out of a vertical cylindrical barrel with base radius \(R\) and altitude \(H.\)

Solution.

Calculating work to pump the water out of a cylindrical barrel with base radius R and altitude H.
Figure 2.

To find the work, we take a thin representative slice with thickness \(dz\) at a height \(z\) from the bottom of the barrel. The mass of the slice is

\[dm = \rho dV = \rho Adz = \pi \rho {R^2}dz.\]

To pump out this volume of water out of the barrel, we need to raise it to the height \(H\). The work required for this is given by the expression

\[dW = dm \cdot g\left( {H - z} \right) = \pi \rho g{R^2}\left( {H - z} \right)dz.\]

The total amount of work is found by integration from \(z = 0\) to \(z = H:\)

\[W = \int\limits_0^H {dW} = \int\limits_0^H {\pi \rho g{R^2}\left( {H - z} \right)dz} = \pi \rho g{R^2}\int\limits_0^H {\left( {H - z} \right)dz} = \pi \rho g{R^2}\left. {\left( {Hz - \frac{{{z^2}}}{2}} \right)} \right|_0^H = \frac{{\pi \rho g{R^2}{H^2}}}{2}.\]

Example 3.

Calculate the work that has to be done to raise a body of mass \(m\) from the Earth's surface to an altitude \(h.\) What is the work if the body is removed to infinity?

Solution.

Assuming the radius of Earth is \(R,\) the mass of Earth is \(M,\) and acceleration due to gravity at its surface is \(g,\) we write the gravitational force acting on the body at the Earth's surface in the form

\[{F_0} = G\frac{{mM}}{{{R^2}}} = mg,\]

where \(G\) is the gravitational constant, and \(g = \frac{{GM}}{{{R^2}}}\) is the acceleration due to gravity.

At a certain height \(x,\) the gravitational force is given by

\[F\left( x \right) = G\frac{{mM}}{{{{\left( {R + x} \right)}^2}}} = G\frac{{mM{R^2}}}{{{{\left( {R + x} \right)}^2}{R^2}}} = \frac{{mg{R^2}}}{{{{\left( {R + x} \right)}^2}}}.\]

The work to raise the body to an altitude \(h\) is determined through integration:

\[W = \int\limits_0^h {F\left( x \right)dx} = \int\limits_0^h {\frac{{mg{R^2}}}{{{{\left( {R + x} \right)}^2}}}dx} = mg{R^2}\int\limits_0^h {\frac{{dx}}{{{{\left( {R + x} \right)}^2}}}} = mg{R^2}\left. {\left( { - \frac{1}{{R + x}}} \right)} \right|_0^h = mg{R^2}\left( { - \frac{1}{{R + h}} + \frac{1}{R}} \right) = mg\frac{{{R^\cancel{2}}h}}{{\cancel{R}\left( {R + h} \right)}} = mg\frac{{Rh}}{{R + h}}.\]

The work needed to move the body from the Earth's surface to infinity is given by the limit

\[{W_\infty } = \lim\limits_{h \to \infty } W = \lim\limits_{h \to \infty } \left( {mg\frac{{Rh}}{{R + h}}} \right) = mg\lim\limits_{h \to \infty } \frac{{Rh}}{{R + h}} = mg\lim\limits_{h \to \infty } \frac{R}{{\frac{R}{h} + 1}} = mg\frac{R}{{0 + 1}} = mgR.\]

Example 4.

A raindrop with initial mass \({m_0}\) starts falling from rest under the action of gravity and evenly evaporates losing every second mass equal to \(\mu.\) What is the work of gravity during the time from the beginning of the movement to the complete evaporation of the drop. Neglect air resistance.

Solution.

The mass of the raindrop varies according to the law

\[m\left( t \right) = {m_0} - \mu t.\]

So the complete evaporation time is

\[T = \frac{{{m_0}}}{\mu }.\]

Determine the elementary work over an infinitesimally small time interval \(\left[ {t,t + dt} \right].\) The force of gravity at the moment \(t\) is

\[F\left( t \right) = m(t)g = \left( {{m_0} - \mu t} \right)g.\]

For the time \(dt,\) the drop moves a distance equal to

\[dx = v(t)dt = gtdt.\]

Integrating from \(t = 0\) to \(t = T = \frac{{{m_0}}}{\mu }\) gives the total work:

\[W = \int\limits_0^T {dW\left( t \right)} = \int\limits_0^T {F\left( t \right)} dx\left( t \right) = \int\limits_0^T {\left( {{m_0} - \mu t} \right){g^2}tdt} = {g^2}\int\limits_0^T {\left( {{m_0}t - \mu {t^2}} \right)dt} = {g^2}\left. {\left( {\frac{{{m_0}{t^2}}}{2} - \frac{{\mu {t^3}}}{3}} \right)} \right|_0^T = {g^2}\left( {\frac{{{m_0}{T^2}}}{2} - \frac{{\mu {T^3}}}{3}} \right) = {g^2}\left( {\frac{{m_0^3}}{{2{\mu ^2}}} - \frac{{\mu m_0^3}}{{3{\mu ^3}}}} \right) = \frac{{{g^2}m_0^3}}{{{\mu ^2}}}\left( {\frac{1}{2} - \frac{1}{3}} \right) = \frac{{{g^2}m_0^3}}{{6{\mu ^2}}}.\]

See more problems on Page 2.

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