Calculating Work by Integration (Calculus): Videos & Practice Problems

Calculating Work by Integration (Calculus) is used when a variable force depends on position, written as \(F(x)\). In that case, work is not found with a constant-force formula; instead, it is the area under a force-versus-position graph between two positions. This is written as \(W=\int_a^b F(x)\\,dx\) .

This approach connects the physical idea of adding many tiny contributions of force over distance to the calculus idea of an integral. To evaluate the work, integrate the given force function and then apply the upper and lower bounds. Polynomial and exponential force functions can both be handled with standard integration rules. When the work found is the net work on an object, it can be related to motion through the work-energy theorem, \(W_{\text{net}}=\Delta K\) , linking position-dependent forces to changes in speed.

Calculating Work by Integrating Variable Forces

When calculating the work done by a variable force, especially when the force depends on position, calculus becomes essential. Unlike constant forces where work is simply force times displacement, a variable force requires integrating the force function over the distance moved. For example, if the force is given as a function of position, such as \(F(x) = 3x^2\), the work done moving an object from position \(x = 2\) meters to \(x = 5\) meters is found by calculating the area under the force versus position curve between these two points.

Since the force varies with position, the graph of \(F(x)\) is curved, making it impossible to find the exact area using simple geometric shapes like rectangles or triangles. Instead, the work done is determined by summing up an infinite number of infinitesimally thin rectangles under the curve, which is the fundamental concept of an integral in calculus. Mathematically, the work \(W\) done by a variable force between two points \(a\) and \(b\( is expressed as:

\[W = \int_a^b F(x) \, dx\]

Applying this to the example where \)F(x) = 3x^2\) and the limits are from \(2\) to \$5\(, the work done is:

\[W = \int_2^5 3x^2 \, dx\]

To solve this integral, increase the exponent by one and divide by the new exponent, following the power rule of integration:

\[\int 3x^2 \, dx = 3 \cdot \frac{x^{3}}{3} = x^{3}\]

Evaluating this from \)2$ to \(5\) involves substituting the upper and lower limits and subtracting:

\[W = 5^{3} - 2^{3} = 125 - 8 = 117 \text{ joules}\]

This result represents the total work done by the variable force over the specified distance. Understanding how to integrate force functions over displacement is crucial for accurately calculating work in scenarios where forces are not constant, reinforcing the connection between calculus and physical concepts like work and energy.

A metal can is pushed from $x=0.5m$ with a force given by the function $F\left(x\right)=2e^{-2x}$. How much work is done by the force on the metal can?

• A. $0.636J$
• B. $0.318J$
• C. $1.214J$
• D. $0.050J$

Answer: (B)

You push a box with a force given by the function $F\left(x\right)=13-0.6x$. Assuming there's no friction forces, and the box is initially at rest at $x=0$, how fast is the box moving after it has traveled a distance of $20m$?

• A. $5.37\text{ m/s}$
• B. $2.83\text{ m/s}$
• C. $7.48\text{ m/s}$
• D. $24.95\text{ m/s}$

Answer: (C)

Here's what students ask on this topic:

How do you calculate work done by a variable force using integration?

To calculate the work done by a variable force, you use the integral of the force function over the displacement interval. When the force depends on position, represented as $F\left(x\right)$, the work done from position $a$ to $b$ is given by the integral $W={\int }_a^{b}F\left(x\right)dx$. This integral represents the area under the force versus position curve between $a$ and $b$. You evaluate this integral by finding the antiderivative of $F\left(x\right)$ and then applying the limits of integration. This method accounts for the changing force over the distance, unlike the constant force formula.

What is the physical meaning of the integral when calculating work done by a variable force?

The integral used to calculate work done by a variable force represents the sum of many infinitesimally small contributions of force applied over tiny displacements. Physically, this means that instead of assuming a constant force, you consider how the force changes at every point along the path. The integral $W={\int }_a^{b}F\left(x\right)dx$ calculates the total work by adding up all these small amounts of work done over each tiny segment of distance. Graphically, this corresponds to finding the area under the force versus position curve between the two points, which gives the total energy transferred by the force.

How do you evaluate the integral of a polynomial force function to find work?

When the force function is a polynomial, such as $F\left(x\right)=\; 3x^2$, you integrate by increasing the power of $x$ by one and dividing by the new power. For example, the integral of $3x^2$ with respect to $x$ is $3\; \backslash (\backslash times\backslash )\; \backslash (\backslash frac\{x^\{3\}\backslash )\}\{3\}\; =\; x^\{3\}$. After finding the antiderivative, you evaluate it at the upper and lower limits of the interval and subtract: $W\; =\; x^\{3\}\; \backslash (\backslash big\backslash )|\_\{a\}^\{b\}\; =\; b^\{3\}\; -\; a^\{3\}$. This gives the total work done by the force between positions $a$ and $b$.

Why can't you use simple geometric shapes to find work done by a variable force?

Simple geometric shapes like rectangles or triangles only work for constant or piecewise linear forces because their areas can be easily calculated. However, when the force varies continuously with position, the force versus position graph is curved. This means the area under the curve, which represents work, cannot be accurately found using basic shapes. Instead, calculus is used to approximate the area by summing many infinitesimally thin rectangles, leading to the integral formula for work. This method provides an exact calculation of work for any variable force function.

How is the work-energy theorem related to calculating work by integration?

The work-energy theorem states that the net work done on an object equals its change in kinetic energy: $W\_\{net\}\; =\; \backslash (\backslash Delta\backslash )\; K$. When the force depends on position, calculating the net work requires integrating the force function over the displacement. By finding $W\; ={\int }_a^{b}F(x)\; dx$, you determine the total work done, which can then be used to find how the object's kinetic energy changes. This links the calculus method of calculating work with the physical concept of energy changes in motion.