7. Work and Kinetic Energy#
Mar 14, 2026 | 6612 words | 33 min read
7.1. Work#
In physics, work isn’t just about what makes you tired or fatigued. It has a special definition, which is when energy is transferred to an object due the work done on it. Work is done when a force acts on something that undergoes a displacement from one position to another.
Forces can vary as a function of position, and displacements can be along various paths between two points. We define the increment of work \(dW\) done by a force \(\vec{F}\) acting through an infinitesimal displacement \(d\vec{r}\) as the dot product of these two vectors:
We can find the total work done by adding up the contributions for infinitesimal displacements (along a path between two positions).
Work Done by a Force
The work done by a force is the integral of the force with respect to the displacement along the path of the displacement:
The path or line integral adds up contribution along a path. Often the path is parameterized by time so there are time dependent functions \(x(t)\) and \(y(t)\) that are separable. In short, we add up all the components of the force that are parallel to direction of the path (as shown in Fig. 7.1)
Fig. 7.1 Image Credit: Openstax#
Although, we expressed the dot product using the product of magnitudes and the cosine of the angle between them. If it is convenient, one could use the form of the dot product that includes the vector components.
Note
Recall that the vector components of force are portions of a force along a given direction. The components can be positive, negative, or zero, depending on whether the force is directed along the path or perpendicular to it. The maximum work is done either parallel (\(\cos{0}=1\)) or antiparallel (\(\cos{\pi} = -1\)), while zero work is done when the force is perpendicular to the displacement (\(\cos{\theta}=0\)).
The units of work are force multiplied by units of length, which in the SI system is \(\rm N\cdot m\). This combination is called a joule and is abbreviated as \(\rm J\). In the English system (used in the US), the unit of work is the foot-pound (\(\rm ft\cdot lb\)) because the unit of force is the pound (\(\rm lb\)) and the displacement is the foot (\(\rm ft\)).
7.1.1. Work Done by Constant Forces and Contact Forces#
The simplest work to evaluate is that done by a constant magnitude force in a non-varying (single) direction. In this case, the work is only determined by the distance traveled.
Work is an integral. Consider a rectangle of height \(h\) and a length \(\ell\). If the length is divided in to many (\(n\)) small chunks of length \(\Delta \ell_n\). You can visualize that the area of the rectangle can be calculated by summing up all smaller areas, i.e., \(A = \sum_n h\times \Delta \ell_n = h\times \ell.\)
Fig. 7.2 Image Credit: Openstax#
Figure 7.2 shows two people exerting a constant force.
In Fig. 7.2a, the man exerts a constant force \(\vec{F}\) along the handle of a lawn mower, where the horizontal displacement is \(\vec{d}\). Thus, the work down on the lawnmower is \(W = \vec{F}\cdot \vec{d} = Fd\cos{\theta}\), where \(\theta\) is the angle relative to the horizontal.
In Figs. 7.2b and 7.2c shows a woman holding a briefcase. The woman must exert an upward force to balance the weight of the briefcase. In both cases the work done is zero because either in (b) displacement is zero or in (c) the force is perpendicular to the displacement.
When you mow the grass, other forces act on the lawn mower (i.e., contact force of the ground and gravitational force). For an object moving on a surface, the displacement \(d\vec{r}\) is tangent (or parallel) to the surface. The normal force \(\vec{N}\) is perpendicular (or normal) to the surface (i.e., \(\theta = 90^\circ\)). As a result, we have
The normal force never does work under these circumstances. The part of the contact force on the object that is parallel to the surface is friction \(\vec{f}.\) For an object sliding along the surface, the kinetic friction \(\vec{f}_k\) is opposite to \(d\vec{r}\) so the work done by kinetic friction is negative. We can write this as an integral by
where \(|\ell_{AB}|\) is the path length on the surface.
The force of static friction does no work (in the reference frame) because in the case of static friction, there is never any displacement.
As an external force, static friction can do work. Static friction can perform
positive work to prevent someone from sliding off a sled when the sled is moving.
negative work (via air resistance) to balance the positive work (via road on the drive wheels) for a car moving down a flat highway.
You need to analyze each particular case to determine the work done by the forces, whether positive, negative, or zero.
The gravitational force (or weight) of the lawn mower has a constant magnitude \(mg\) and direction, vertically down. The work done by gravity on the mower is the dot product of its weight and its displacement. It is convenient to express the \(x\)-axis along the horizontal and the \(y\)-axis vertically up. Then the gravitational force is \(-mg\hat{j}\), so the work done by gravity \(W_g\) over any path \(A\) to \(B\) is
The work done on an object by gravity depends only on the object’s weight (\(mg\)) and the difference in height \(h = y_B - y_A\) the object is displaced.
Gravity does negative work that moves upward (i.e., \(y_B > y_A\)), which means you must do positive work (against gravity) to lift an object upward.
Alternatively, gravity does positive work on an object that moves downward (\(y_B< y_A\)), which means that you do negative work against gravity to control the descent of the object so it doesn’t drop to the ground.
7.1.1.1. Example Problem: Work to Push a Lawn Mower#
Exercise 7.1
The Problem
How much work is done on the lawn mower by the person in Figure 7.2 (a) if he exerts a constant force of \(75\ \mathrm{N}\) at an angle \(35^\circ\) below the horizontal and pushes the mower \(25\ \mathrm{m}\) on level ground?
The Model
The lawn mower moves on level ground, so its displacement is purely horizontal. The applied force has a constant magnitude and direction, making a fixed angle \(\theta\) with the displacement. Only the component of the applied force parallel to the displacement contributes to the work done on the mower. Because both the force and displacement are constant, the definition of work for a constant force applies.
The Math
The work done by a constant force is defined as the dot product of the force and the displacement,
Here, \(F\) is the magnitude of the applied force, \(d\) is the magnitude of the displacement, and \(\theta\) is the angle between them.
Substituting the given values, we find
The Conclusion
The work done on the lawn mower by the applied force is \(\boxed{W = 1500\ \mathrm{J}}.\) The result is reported with appropriate significant figures based on the given quantities.
The Verification
We verify the analytic result by computing the work numerically using Python.
import numpy as np
F = 75.0 # N
d = 25.0 # m
theta = np.deg2rad(35.0)
W = F * d * np.cos(theta)
print("The work done on the lawn mower is %d N." % np.round(W,-2)) #The -2 forces it to round to the hundreds place.
7.1.1.2. Example Problem: Moving a Couch#
Exercise 7.2
The Problem
You decide to move your couch to a new position on your horizontal living room floor. The normal force on the couch is \(1\ \mathrm{kN}\) and the coefficient of friction is \(0.6\). (a) You first push the couch \(3\ \mathrm{m}\) parallel to a wall and then \(1\ \mathrm{m}\) perpendicular to the wall (\(A\) to \(B\) in Figure 7.3). How much work is done by the frictional force? (b) You don’t like the new position, so you move the couch straight back to its original position (\(B\) to \(A\) in Figure 7.3). What was the total work done against friction moving the couch away from its original position and back again?
Fig. 7.3 Image Credit: Openstax#
The Model
The couch moves on a horizontal surface, so the normal force is constant. The kinetic friction force has constant magnitude and always opposes the direction of motion. The magnitude of the kinetic friction force is given by \(f_k = \mu_k N\). The work done by friction is negative because the friction force is opposite the displacement. Path length, not displacement, determines the work done by friction.
The Math
The magnitude of the kinetic friction force is
Substituting the given values,
(a) The total path length from \(A\) to \(B\) is the sum of the two straight segments,
The work done by friction is therefore
Substituting values,
(b) The return path from \(B\) to \(A\) is the hypotenuse of a right triangle with legs \(3\ \mathrm{m}\) and \(1\ \mathrm{m}\), so its length is
The total path length for the round trip is
The total work done against friction is the negative of the work done by friction,
Substituting values,
Evaluating this expression gives
The Conclusion
The work done by friction in moving the couch from \(A\) to \(B\) is \(\boxed{W_{\mathrm{fr}} = -2400\ \mathrm{J}}.\) The total work done against friction in moving the couch away from its original position and back again is \(\boxed{W_{\text{against}} = 4300\ \mathrm{J}}.\)
This result shows that friction is a nonconservative force, since the total work depends on the path taken even when the initial and final positions are the same.
The Verification
We verify the total work done against friction by computing the friction force and total path length numerically.
import numpy as np
mu = 0.6
N = 1.0e3 # N
fk = mu * N
d_total = 3 + 1 + np.sqrt(10)
W_against = fk * d_total
print("The total work done against friction is %.1e J." % W_against)
7.1.1.3. Example Problem: Shelving a Book#
Exercise 7.3
The Problem
You lift an oversized library book, weighing \(20\ \mathrm{N}\), \(1\ \mathrm{m}\) vertically down from a shelf, and carry it \(3\ \mathrm{m}\) horizontally to a table (Figure 7.4). (a) How much work does gravity do on the book? (b) When you’re finished, you move the book in a straight line back to its original place on the shelf. What was the total work done against gravity, moving the book away from its original position on the shelf and back again?
Fig. 7.4 Image Credit: Openstax#
The Model
The book moves in a vertical plane under the influence of gravity. Gravity acts with constant magnitude and direction throughout the motion. The work done by gravity depends only on the change in vertical position, not on the path taken. Horizontal motion contributes no work by gravity because the force of gravity is vertical.
The Math
The work done by gravity for vertical motion is given by
where \(W\) is the weight of the book and \(y\) is the vertical position measured upward.
(a) The book is moved downward by \(1\ \mathrm{m}\), so
Substituting into the work expression, we find
(b) The book begins and ends at the same vertical position on the shelf, so the net change in height is
Therefore, the total work done by gravity over the closed path is \(W_g = 0.\)
Because work done against gravity is the negative of the work done by gravity, the total work done against gravity is also zero.
The Conclusion
The work done by gravity while the book is moved from the shelf to the table is \(\boxed{W_g = 20\ \mathrm{J}}.\) The total work done against gravity in moving the book away from the shelf and back again is \(\boxed{0\ \mathrm{J}}.\)
This result demonstrates that gravity is a conservative force, since the total work over any closed path is zero.
The Verification
We verify the result by computing the work done by gravity using the change in vertical position.
W = 20.0 # N (weight of the book)
dy = -1.0 # m (downward displacement)
Wg = -W * dy
print("The work done by gravity during the downward motion is %d J." % Wg)
print("The total work done against gravity over the closed path is 0 J.")
7.1.2. Work Done by Forces that Vary#
Forces may vary in magnitude and direction at different points in space. Additionally the paths between two points can be curved rather than straight. The infinitesimal work done by a variable force can be expressed in terms of the force components and the displacement along the path,
Here the force is assumed to be a function of the position, for example,
and the displacements depend on the equations on the path, or
Equation (7.2) defines the total work as a line integral, which sums the infinitesimal amounts of work along the path. The physical concept of work is straightforward: (1) you calculate the work for tiny displacements and (2) add them up.
One very important and widely applicable variable force is the describe by Hooke’s law (i.e., spring force),
where \(k\) is the spring constant, and \(\Delta\vec{x} = \vec{x}-\vec{x}_{\rm eq}\) is the displacement from the spring’s unstretched (equilibrium) position. Forces between molecules (or any system undergoing displacements from a stable equilibrium) behave approximately like a spring force.
Fig. 7.5 Image Credit: Openstax#
To calculate the work done by a spring force, we can choose the \(x\)-axis along the length of the spring (see Fig. 7.5).
As the length increases, the end position of the spring moves to the right (\(\hat{i}\)) direction, where the equilibrium point \(x_{\rm eq} = 0\).
Conversely as the length decreases (i.e., spring compresses), the end position of the spring moves to the left (\(-\hat{i}\)).
With this choice of coordinates, the spring force has only an \(x\)-component, \(F_x = -kx\), and the work done when \(x\) changes from \(x_A\) to \(x_B\) is
Work Done by a Spring Force
The work done by the spring \(W_{AB}\) depends only on the starting and ending points, \(A\) and \(B\). It is independent of the actual path between them, as long as it starts at \(A\) and ends at \(B\). The actual path could involve going back and forth before ending.
For the 1D case involving the work done by a spring force, you can readily see the correspondence between the work done by a force and the area under the curve of the force versus its displacement. In general, a 1D integral is the limit of the sum of infinitesimals (\(f(x)dx\)) represented by rectangles (strips) as shown in Figure 7.6.
Fig. 7.6 The area under any continuous curve can be approximated by drawing a number of rectangles. The integral is the limit for an infinite number of rectangles. Figure Credit: Hyperphysics#
Since \(F=-kx\) represents a straight line with slope \(-k\), the area under the curve can be calculated by two triangles, where one area is positive (above the \(x\)-axis) and the other is negative (below the \(x\)-axis). The magnitude of one of these “areas” is just one-half the triangle’s base times the triangle’s height (e.g., \(A = bh/2\)) along the force axis. See the code below and the figure it produces. You verify the areas using simple geometry (i.e., areas of triangles and trapezoids).
import numpy as np
import matplotlib.pyplot as plt
fs = 'large'
k = 1 #spring constant in N/m
x_rng = np.arange(-3,3.1,0.1) #range of positions
def spring_force(x,k):
return -k*x
fig = plt.figure(figsize=(5,5),dpi=120)
ax = fig.add_subplot(111)
ax.grid(True,alpha=0.1,color='k')
#Plot force line
ax.plot(x_rng,spring_force(x_rng,k),'k--',lw=2)
x_left = np.arange(-1.5,0.1,0.1)
#fill red area
ax.fill_between(x_left,spring_force(x_left,k),color='r',alpha=0.25)
#fill cyan/blue area
x_right1,x_right2 = np.arange(0,1.6,0.1), np.arange(1.5,3.1,0.1)
ax.fill_between(x_right1,spring_force(x_right1,k),color='c',alpha=0.45)
ax.fill_between(x_right2,spring_force(x_right2,k),color='darkblue',alpha=0.25)
#annotate x-limits
ax.text(-1.5,-0.35,'$-x_A$',color='r',horizontalalignment='center',fontsize=fs)
ax.text(1.5,0.15,'$x_A$',color='c',horizontalalignment='center',fontsize=fs)
ax.text(3,0.15,'$x_B$',color='darkblue',horizontalalignment='center',fontsize=fs)
# ---- annotations ----
# f(x) = -kx label with curved arrow
ax.annotate(r'$F(x)=-kx$',xy=(-0.7, 0.7),xytext=(0.4, 1.8),
fontsize=fs,arrowprops=dict(arrowstyle='->',lw=1.5,connectionstyle='arc3,rad=-0.25'))
# Positive area annotation (left, red)
ax.annotate('Positive\narea',xy=(-1.1, 0.6),xytext=(-2.4, 1.1),
fontsize=fs,ha='center',arrowprops=dict(arrowstyle='->',lw=1.5,connectionstyle='arc3,rad=0.3'))
# Negative areas annotation (right, blue)
ax.annotate('Negative\nareas',xy=(1.2, -0.9),xytext=(1, -2.5),
fontsize=fs,ha='center',arrowprops=dict(arrowstyle='->',lw=1.5,connectionstyle='arc3,rad=-0.3'))
ax.annotate('',xy=(2.2, -0.9),xytext=(1.5, -2.5),
fontsize=fs,ha='center',arrowprops=dict(arrowstyle='->',lw=1.5,connectionstyle='arc3,rad=0.3'))
# ---- spine-centered axes ----
ax.spines['left'].set_position('zero')
ax.spines['bottom'].set_position('zero')
ax.spines['right'].set_color('none')
ax.spines['top'].set_color('none')
ax.xaxis.set_ticks_position('bottom')
ax.yaxis.set_ticks_position('left')
#set plot limits
ax.set_xlim(-3.5,3.5)
ax.set_ylim(-3.5,3.5);
7.1.2.1. Example Problem: Work Done by a Variable Force#
Exercise 7.4
The Problem
An object moves along a parabolic path \(y = (0.5\ \mathrm{m^{-1}})x^2\) from the origin \(A = (0,0)\) to the point \(B = (2\ \mathrm{m}, 2\ \mathrm{m})\) under the action of a force \(\vec{F} = (5\ \mathrm{N/m})y\,\hat{i} + (10\ \mathrm{N/m})x\,\hat{j}\) (Figure 7.7). Calculate the work done.
Fig. 7.7 Image Credit: Openstax#
The Model
The object moves in the \(x\)–\(y\) plane along a specified curved path. The force varies with position and has both \(x\)- and \(y\)-components. The work done by a variable force is calculated using a line integral of \(\vec{F} \cdot d\vec{r}\) along the path. The path is described parametrically using \(x\) as the independent variable.
The Math
The infinitesimal work done by a force is
The equation of the path gives
Differentiating with respect to \(x\),
Substituting the force components and the path relations into the expression for \(dW\) gives
The total work is obtained by integrating from \(x=0\) to \(x=2\ \mathrm{m}\),
Evaluating the integral, we find
The Conclusion
The work done by the force as the object moves along the parabolic path is \(\boxed{W = 33\ \mathrm{J}}.\) The result is reported with appropriate significant figures based on the given quantities.
The Verification
We verify the analytic result by approximating the integral as a sum of many small trapezoids of width \(\Delta x\). We then compare this manual trapezoid sum to np.trapz using the same \(\Delta x\), and confirm the results agree within a small tolerance.
import numpy as np
import matplotlib.pyplot as plt
# -----------------------
# Integrand setup
# -----------------------
def integrand(x):
return 12.5 * x**2
a, b = 0.0, 2.0
# Choose a small step size for the numerical check
dx = 0.01
x = np.arange(a, b + dx, dx)
y = integrand(x)
# -----------------------
# Manual trapezoid sum (sum of discrete trapezoids)
# -----------------------
W_manual = 0.0
for i in range(len(x) - 1):
W_manual += 0.5 * (y[i] + y[i + 1]) * dx
# -----------------------
# Compare to NumPy trapezoid method (should match)
# -----------------------
W_trapz = np.trapz(y, x)
diff = abs(W_manual - W_trapz)
print("Using Δx = %.2f m:" % dx)
print("Manual trapezoid sum gives W = %.3f J." % W_manual)
print("np.trapz gives W = %.3f J." % W_trapz)
print("The absolute difference is %.6f J." % diff)
# A simple tolerance that scales with dx (tight enough for this smooth function)
tol = 1e-6
print("The two methods agree within the tolerance of %.1e J: %s." % (tol, diff < tol))
# -----------------------
# Graphical illustration: show trapezoids (use a coarser dx for clarity)
# -----------------------
dx_plot = 0.25
xp = np.arange(a, b + dx_plot, dx_plot)
yp = integrand(xp)
# Plot the smooth curve on a fine grid for context
xfine = np.arange(a, b, 0.01)
yfine = integrand(xfine)
fig = plt.figure(figsize=(6,4), dpi=120)
ax = fig.add_subplot(111)
ax.plot(xfine, yfine, lw=2)
# Draw trapezoids under the curve using the coarse grid
for i in range(len(xp) - 1):
x0, x1 = xp[i], xp[i + 1]
y0, y1 = yp[i], yp[i + 1]
ax.fill([x0, x0, x1, x1], [0, y0, y1, 0], alpha=0.25)
ax.set_xlabel("x (m)")
ax.set_ylabel("integrand (${\\rm J/m}$)")
ax.grid(True, alpha=0.2)
ax.set_xlim(0,2)
ax.set_ylim(0,50)
plt.show()
Using Δx = 0.01 m:
Manual trapezoid sum gives W = 33.334 J.
np.trapz gives W = 33.334 J.
The absolute difference is 0.000000 J.
The two methods agree within the tolerance of 1.0e-06 J: True.
7.1.2.2. Example Problem: Work Done by a Spring Force#
Exercise 7.5
The Problem
A perfectly elastic spring requires \(0.54\ \mathrm{J}\) of work to stretch \(6\ \mathrm{cm}\) from its equilibrium position, as in Figure 7.5(b). (a) What is its spring constant \(k\)? (b) How much work is required to stretch it an additional \(6\ \mathrm{cm}\)?
The Model
The spring obeys Hooke’s law, so the force exerted by the spring is proportional to the displacement from equilibrium. The work required to stretch the spring is the work done against the spring force. Because the spring force is conservative, the work depends only on the initial and final displacements, not on the path taken. The displacement \(x\) is measured from the equilibrium position.
The Math
The work required to stretch a spring from position \(x_A\) to \(x_B\) is
(a) For the initial stretch, the spring moves from equilibrium to \(x_B = 6\ \mathrm{cm}\), so \(x_A = 0\).
Substituting the given values and solving for the spring constant gives,
(b) For the additional stretch, the spring moves from \(x_A = 6\ \mathrm{cm}\) to \(x_B = 12\ \mathrm{cm}\).
Using the value of \(k\) found in part (a),
The Conclusion
The spring constant is \(\boxed{k = 3\ \mathrm{N/cm}}.\) The work required to stretch the spring an additional \(6\ \mathrm{cm}\) is \(\boxed{W = 1.6\ \mathrm{J}}.\) This result reflects the quadratic dependence of spring work on displacement from equilibrium.
The Verification
We verify the results by directly evaluating the spring-work expression using Python.
# Given values
W1 = 0.54 # J
x1 = 6.0 # cm
# Part (a): solve for k
k = 2 * W1 / (x1**2)
# Part (b): compute additional work from 6 cm to 12 cm
x2 = 12.0 # cm
W2 = 0.5 * k * (x2**2 - x1**2)
print("The spring constant is %.1f N/cm." % k)
print("The work required to stretch the spring from 6 cm to 12 cm is %.1f J." % W2)
7.2. Kinetic Energy#
It’s plausible to suppose that a larger velocity of a body will have a greater effect on other bodies. This does not depend on the direction of the velocity only its magnitude (i.e., speed). The larger effect can be translated into a larger potential to do work once the fast moving body strikes a slower one.
An example is through billiards (or pool), when the fast moving (white) cue ball strikes the blue-colored 2 ball. The faster the cue, the more the 2 ball is displaced. The cue ball must have some “energy” that is transferred to the 2 ball. The cue must have some “energy of motion” as its traveling across the table. During the 18th century (i.e., 1700s), the “energy of motion” was renamed to kinetic energy.
The kinetic energy has two definitions: (1) classical and (2) relativistic. The relativistic definition only pertains to objects moving near the speed of light, which we are not investigating here. We’ll stick with the classical definition.
Kinetic Energy
The kinetic energy \(K\) of a particle is one-half the product of the particle’s mass \(m\) and the square of its speed \(v\). We extend this definition to any system of particles by adding up the kinetic energies of all the constituent particles to get
Recall from Newton’s 2nd Law and Momentum, we represented Newton’s 2nd law in a different form using the momentum \(\vec{p} = m\vec{v}\). We can do the same here using the magnitude \(p = \sqrt{\vec{p}\cdot \vec{p}}= mv\), where the kinetic energy becomes
which is sometimes more convenient to use.
The units of kinetic energy are the mass times the speed squared, or \(\rm kg\cdot m^2/s^2\). Recalling that the units for work are \(\rm N\cdot m = kg\cdot m/s^2\cdot m = kg\cdot m^2/s^2\), which are joules (\(\rm J\)). Work and kinetic energy have the same units because they are different forms of the same (more general) physical property.
Because velocity is a relative quantity, then the kinetic energy must depend on your frame of reference. A frame of reference is chosen to simplify your calculations and subsequent analysis.
One frame of reference is external to the system (i.e., where the observations are made).
Another frame lies attached to (or moves with) the system as an internal frame.
The equations for relative motion provide a link to calculating the kinetic energy of an object with respect to different frames of reference. For example,
represents the kinetic energy of an object with speed \(v_A\) that is moving relative to another object \(v_B\). Depending on the relative direction of motion, we can determine whether there is a \(+\) or \(-\).
7.2.1. Example Problem: Kinetic Energy of an Object#
Exercise 7.6
The Problem
(a) What is the kinetic energy of an \(80\ \mathrm{kg}\) athlete, running at \(10\ \mathrm{m/s}\)?
(b) The Chicxulub crater in Yucatán, one of the largest existing impact craters on Earth, is thought to have been created by an asteroid traveling at \(22\ \mathrm{km/s}\) and releasing \(4.2\times10^{23}\ \mathrm{J}\) of kinetic energy upon impact. What was its mass?
(c) In nuclear reactors, thermal neutrons, traveling at about \(2.2\ \mathrm{km/s}\), play an important role. What is the kinetic energy of such a particle?
The Model
Each object is treated as a point mass moving at a nonrelativistic speed, so the classical expression for kinetic energy applies. The kinetic energy depends only on the mass and speed of the object and is independent of direction. All quantities are expressed in SI units, so speeds given in kilometers per second must be converted to meters per second before substitution.
The Math
The kinetic energy of an object of mass \(m\) moving at speed \(v\) is defined as
(a) The athlete has mass \(m = 80\ \mathrm{kg}\) and speed \(v = 10\ \mathrm{m/s}\). We substitute these values into the kinetic energy expression, which gives the value of the athlete’s kinetic energy.
(b) The asteroid releases kinetic energy \(K = 4.2\times10^{23}\ \mathrm{J}\) and travels at \(v = 22\ \mathrm{km/s} = 2.2\times10^{4}\ \mathrm{m/s}\). We solve the kinetic energy expression for the mass and substitute the given values, which yields the mass of the impactor by
(c) A thermal neutron has mass \(m = 1.68\times10^{-27}\ \mathrm{kg}\) and speed \(v = 2.2\ \mathrm{km/s} = 2200\ \mathrm{m/s}\). We substitute these values into the kinetic energy expression, which gives the kinetic energy of the neutron as
The Conclusion
Using the definition of kinetic energy, we determined that the athlete running at everyday speeds has a kinetic energy of \(4000\ \mathrm{J}\), while the Chicxulub asteroid carries an enormous kinetic energy of \(4.2\times10^{23}\ \mathrm{J}\), despite both systems being described by the same equation. In contrast, the thermal neutron was found to have a kinetic energy of \(4.1\times10^{-21}\ \mathrm{J}\), reflecting the effect of its extremely small mass even at relatively high speeds. Together, these results reinforce the quadratic dependence of kinetic energy on speed and show how the same physical principle spans an extraordinary range of physical scales, from human motion to planetary impacts and nuclear processes.
The Verification
We verify the results by organizing the data into lists and using a single index-based loop. This shows how the same physical relationship can be applied systematically without repeating code.
import numpy as np
# -----------------------
# Store problem data
# -----------------------
labels = ["athlete", "Chicxulub asteroid", "thermal neutron"]
masses = [80.0, None, 1.68e-27] # Masses (kg); use None if mass is unknown
speeds = [10.0, 2.2e4, 2200.0] # Speeds (m/s)
energies = [None, 4.2e23, None] # Kinetic energies (J); use None if energy is unknown
# -----------------------
# Loop using an index
# -----------------------
for i in range(len(labels)):
label,m,v,K = labels[i], masses[i], speeds[i], energies[i]
if m is not None: # If the mass is known, compute kinetic energy
K_calc = 0.5 * m * v**2
if label == "athlete":
print("The kinetic energy of the athlete is %d J." % K_calc)
else:
print("The kinetic energy of the %s is %.1e J." % (label,K_calc))
else: # If the kinetic energy is known, compute mass
m_calc = 2 * K / v**2
print("The mass of the Chicxulub asteroid is %.2e kg." % m_calc)
7.2.2. Example Problem: Kinetic Energy Relative to Different Frames#
Exercise 7.7
The Problem
A \(75\ \mathrm{kg}\) person walks down the central aisle of a subway car at a speed of \(1.5\ \mathrm{m/s}\) relative to the car, whereas the train is moving at \(15\ \mathrm{m/s}\) relative to the tracks.
(a) What is the person’s kinetic energy relative to the car?
(b) What is the person’s kinetic energy relative to the tracks?
(c) What is the person’s kinetic energy relative to a frame moving with the person?
The Model
The person is treated as a point mass moving at nonrelativistic speeds, so the classical expression for kinetic energy applies. Motion is confined to a single horizontal line, allowing velocities to be treated as signed scalars along one dimension. Different frames of reference measure different speeds for the same object, but each frame is inertial, and kinetic energy is computed using the speed measured in that frame.
The Math
The kinetic energy of an object of mass \(m\) moving at speed \(v\) is defined as
(a) Relative to the subway car, the person moves at a speed of \(v = 1.5\ \mathrm{m/s}\). We substitute this speed and the given mass into the kinetic energy expression, which yields the kinetic energy measured in the car’s frame.
(b) Relative to the tracks, the person’s speed depends on the direction of walking. Because velocities add along the same line, the speed relative to the tracks is given by \(v = 15\ \mathrm{m/s} \pm 1.5\ \mathrm{m/s}\). We evaluate the kinetic energy for each possible speed to determine the values measured in the track frame.
For walking toward the front of the car,
For walking toward the back of the car,
(c) In a frame moving with the person, the person’s speed is zero. Because kinetic energy depends on the square of the speed, a speed of zero implies zero kinetic energy in that frame. \(K = 0.\)
The Conclusion
Relative to the subway car, the person has a modest kinetic energy of \(84.4\ \mathrm{J}\), reflecting the low walking speed inside the car. Relative to the tracks, the kinetic energy is much larger and depends on whether the person walks with or against the train’s motion, ranging from \(6830\ \mathrm{J}\) to \(1.02\times10^{4}\ \mathrm{J}\). In the person’s own rest frame, the kinetic energy is zero. These results show that kinetic energy is not an intrinsic property of an object alone but depends on the observer’s frame of reference, reinforcing the earlier discussion that speed, and therefore kinetic energy, is frame-dependent even in classical mechanics.
The Verification
We verify the results by computing the kinetic energies in different frames using lists and an index-based loop.
import numpy as np
# -----------------------
# Store problem data
# -----------------------
labels = ["car frame", "track frame (forward)", "track frame (backward)", "person frame"]
masses = [75.0, 75.0, 75.0, 75.0] # kg
speeds = [1.5, 16.5, 13.5, 0.0] # m/s
# -----------------------
# Loop using an index
# -----------------------
for i in range(len(labels)):
label, m, v = labels[i], masses[i], speeds[i]
K = 0.5 * m * v**2
if K >= 1e4:
print("The kinetic energy in the %s is %.2e J." % (label, K))
else:
print("The kinetic energy in the %s is %.1f J." % (label, K))
7.3. Work-Energy Theorem#
According to Newton’s 2nd law, the net force (all forces acting on a particle) determines the rate of change in the momentum of the particle or its motion. Therefore, we should consider the work done by all the forces acting on a particle, or the net work, to see what effect it has on the particle’s motion.
Consider the net work done on a particle that moves over an infinitesimal displacement, or
Newton’s 2nd law tells us that we can transform the net force \(F_{\rm net}\) using the acceleration \(d\vec{\rm v}{dt}\), so
Using the definition of velocity, \(d\vec{r} = \vec{\rm v}dt\), we have:
where also used the commutative property of the dot product.
We can represent the dot product in terms of Cartesian coordinates and then include the result in the integral between two points \(A\) and \(B\) on the particle’s trajectory. This gives us the work done on the particle as:
We used the relation for \(v^2 = v_x^2 + v_y^2 + v_z^2\) to reduce the expression and the definition of a particle’s kinetic energy \(K\). The result is called the work-energy theorem.
Work-Energy Theorem
The net work on a particle equals the change in the particle’s kinetic energy:
According to this theorem, when an object
slows down (\(v_f<v_i\)), its final kinetic energy is less than the initial and the net work done on it is negative.
speeds up (\(v_f>v_i\)), its final kinetic energy is greater than the initial and the net work on it is positive.
When calculating the net work, you must include all the forces acting on an object. If you leave out any forces or include extra forces not acting on the object, you will get a wrong result.
The importance of the work-energy theorem is that it makes some calculations much simpler to accomplish compared to solving them with Newton’s 2nd law. For example, we obtained the acceleration of a block on a frictionless incline as \(a = g\sin{\theta}\) using Newton’s second law in Chapter 5 (Example Problem: Weight on an incline). If we also use the kinematic equation for acceleration
then we could find that
where \(s_f\) and \(s_i\) are the initial and final positions on the incline, respectively.
We can also get this result from the work-energy theorem in Equation (7.8). Since only two forces are acting on the object (gravity and the normal force; normal force doesn’t do work), the net work is just the work done by gravity. The work \(dW\) is the dot product of gravity (\(\vec{F}_g = -mg\hat{j}\)) and the displacement (\(\vec{r} = dx\hat{i} + dy\hat{j}\)). Applying the work-energy theorem, we have
We have found the left-hand side of the work-energy theorem, where we now set it equal to the change in kinetic energy:
In Example Problem: Weight on an incline the change in height (\(y_f-y_i\)) forms the short side of a right triangle with the same angle \(\theta\) from the incline. The hypotenuse of the triangle is (\(s_f-s_i\)) along the incline. Therefore, we can show that the short side of the triangle is \((y_f - y_i) = -(s_f-s_i)\sin{\theta}\) using our trigonometric identity for \(\sin{\theta}\). Note: \(s\) is defined to increase down the incline (to the left in the figure), which introduces the minus sign. We can then obtain
What is gained by using the work-energy theorem?
For a frictionless plane, not much. Newton’s 2nd law is easy to solve only for this particular case, whereas the work-energy theorem gives the final speed for any shaped frictionless surface. For an arbitrary curved surface, the normal is not constant, where it may be difficult or impossible to solve analytically using Newton’s 2nd law.
For motion along a surface, the normal force never does any work because it’s perpendicular to the displacement. A calculation using the work-energy theorem avoids this difficulty and applies to more general solutions.
Problem Solving Strategy
Work-Energy Theorem
Draw a free-body diagram for each force on the object.
Determine whether or not each force does work over the displacement in the diagram. Be sure to keep any \(+\) or \(-\) in the work done.
Add up the total amounts of work done by each force.
Set this total work \(W\) equal to the change in kinetic energy \(\Delta K\) and solve for any unknown parameter.
Check your answers.
If the object is traveling at a constant speed or zero acceleration, the total work should be zero and match the change in kinetic energy.
If the total work is positive, the object must have sped up or increased its kinetic energy.
If the total work is negative, the object must have slowed down or decreased its kinetic energy.
7.3.1. Example Problem: Loop-the-Loop#
Exercise 7.8
The Problem
The frictionless track for a toy car includes a loop-the-loop of radius \(R\). How high, measured from the bottom of the loop, must the car be placed to start from rest on the approaching section of track and go all the way around the loop?
Fig. 7.8 Image Credit: Openstax#
The Model
The car is treated as a point mass moving on a frictionless track, so mechanical energy is conserved. Motion occurs in a vertical plane under the influence of gravity and the normal force from the track. The normal force constrains the motion but does no work, so gravity is the only force that changes the car’s mechanical energy. The car starts from rest at height \(y_1\) and must maintain contact with the track at the top of the loop, which imposes a centripetal acceleration requirement.
The Math
The car starts from rest, so its initial kinetic energy is zero. Because gravity is the only force that does work, the work–energy theorem relates the change in gravitational potential energy to the kinetic energy at the top of the loop.
For motion from height \(y_1\) to the top of the loop at height \(y_2 = 2R\), the work–energy theorem gives
At the top of the loop, the car moves in a circle of radius \(R\), so it requires a centripetal acceleration directed downward. The forces acting on the car at this point are gravity and the normal force, both directed toward the center of the loop. Applying Newton’s second law in the radial direction gives
To maintain contact with the track, the normal force must be positive, so the limiting case occurs when \(N = 0\). This condition determines the minimum speed required at the top of the loop. Solving for \(v_2^2\) yields
We substitute this result into the work–energy equation and solve for the required starting height by
The Conclusion
We find that the car must start from a minimum height of \(y_1 = \tfrac{5R}{2}\) above the bottom of the loop in order to complete the loop without losing contact with the track. This height ensures that the car has just enough speed at the top of the loop to provide the required centripetal acceleration when the normal force is reduced to zero. The result demonstrates how the work–energy theorem can be combined with a force condition at a critical point to determine motion without solving differential equations, reinforcing the earlier idea that energy methods often provide the most efficient path to a solution.
The Verification
We verify the result symbolically by checking that the required speed at the top of the loop corresponds to zero normal force and that the work–energy relation yields the same starting height.
import numpy as np
# Choose an arbitrary radius
R = 1.0 # m
g = 9.81 # m/s^2
# Speed at the top from centripetal condition
v2 = np.sqrt(g * R)
# Starting height from work-energy
y1 = 2*R + v2**2/(2*g)
print("The minimum starting height is %.2f R." % (y1 / R))
7.3.2. Example Problem: Determining a Stopping Force#
Exercise 7.9
The Problem
A bullet has a mass of \(40\) grains \((2.6\ \mathrm{g})\) and a muzzle velocity of \(1100\ \mathrm{ft/s}\) \((335\ \mathrm{m/s})\). It can penetrate eight \(1\)-inch pine boards, each with thickness \(0.75\) inches. What is the average stopping force exerted by the wood?
Fig. 7.9 Image Credit: Openstax#
The Model
The bullet is treated as a point mass moving in a straight line through the boards. The interaction with the wood is modeled as a constant average stopping force acting opposite the direction of motion over the total penetration distance. Gravity and air resistance are neglected because their contributions to the work over the short stopping distance are negligible. The bullet is assumed to come to rest after penetrating the boards, so all of its initial kinetic energy is removed by the work done by the stopping force.
The Math
Because the bullet comes to rest, the change in kinetic energy is equal in magnitude to the initial kinetic energy. The work–energy theorem relates this change in kinetic energy to the work done by the average stopping force.
The total stopping distance is the combined thickness of the boards. Each board has thickness \(0.75\ \mathrm{in}\), so the total penetration distance is
The work done by the stopping force is negative because the force opposes the motion, so the work–energy theorem gives
Solving this expression for the magnitude of the average stopping force yields
We substitute the given mass, speed, and stopping distance and evaluate.
The Conclusion
We find that the average stopping force exerted by the wood on the bullet is \(960\ \mathrm{N}\). This force represents the constant force that would remove the bullet’s entire initial kinetic energy over the total penetration distance of the boards. The result illustrates how a relatively modest penetration distance can correspond to a large stopping force when an object begins with a high speed, reinforcing the earlier discussion that work–energy methods provide a direct way to relate forces, distances, and changes in kinetic energy without solving equations of motion.
The Verification
We verify the result by recomputing the bullet’s initial kinetic energy and dividing by the stopping distance to obtain the average stopping force.
import numpy as np
# Given values
m = 2.6e-3 # kg
v = 335.0 # m/s
s = 0.152 # m
# Kinetic energy and average stopping force
K = 0.5 * m * v**2
F_ave = K / s
print("The average stopping force exerted by the wood is %.0f N." % F_ave)
7.4. Power#
Thus far, we have not included how work or energy can change over time. We express the relation between work done and the time interval in doing it by introducing the concept of power. Since work can vary as a function of time, we first define average power as the work done \(\Delta W\) during a time interval \(\Delta t\), or
We define the instantaneous power as the work done \(dW\) over an infinitesimal time \(dt\).
Power
Power is defined as the rate of doing work, or the limit of the average power for time intervals approaching zero,
If the power is constant over a time interval, the average power for that interval equals the instantaneous power, and the work done by the agent supplying the power is determined as \(W = P\Delta t\). If the power during the interval varies with time, then the work is the time integral of the power,
The work-energy theorem relates how work can be transformed into kinetic energy. We can also define power as the rate of transfer of energy. Work and energy are measured in units of joules (\(\rm J\)), so power is measured in units of joules per second (\(\rm J/s\)), which has been given the SI name Watts (\(1\ {\rm J/s} = 1\ {\rm W}\)). Another common unit for expressing power capability of everyday devices is horsepower: \(1\ {\rm hp} = 746\ {\rm W}\).
The power involved in moving a body can also be expressed in terms of the forces acting on it. If a force \(\vec{F}\) acts on a body that is displaced by \(d\vec{r}\) in a time \(dt\), the power expended by the force is
where \(\vec{\rm v}\) is the velocity of the body.
7.4.1. Example Problem: Pull-Up Power#
Exercise 7.10
The Problem
An \(80\ \mathrm{kg}\) army trainee does pull-ups on a horizontal bar. It takes the trainee \(0.8\ \mathrm{s}\) to raise the body from a lower position to where his chin is above the bar. How much power do the trainee’s muscles supply moving the body from the lower position to where his chin is above the bar? (Hint: Make reasonable estimates for any quantities needed.)
Fig. 7.10 Image Credit: Openstax#
The Model
The trainee is treated as a point mass moving vertically upward against gravity. The motion is assumed to occur at roughly constant speed, so changes in kinetic energy are neglected and the muscles primarily do work against gravity. The vertical displacement is estimated as \(\Delta y = 0.60\ \mathrm{m}\), corresponding to a typical pull-up height. The arms are assumed to make up \(10\%\) of the trainee’s mass and are not lifted through the full distance, so only \(90\%\) of the body mass is treated as being raised.
The Math
When an object is lifted vertically at constant speed, the work done against gravity over a vertical displacement \(\Delta y\) is given by
Power is defined as the rate at which work is done, so the average power supplied by the muscles is
We substitute the effective mass, gravitational acceleration, vertical displacement, and time and evaluate.
The Conclusion
We find that the trainee’s muscles supply an average power of \(530\ \mathrm{W}\) while performing the pull-up. This power represents the rate at which chemical energy in the muscles is converted into gravitational potential energy of the body. The result connects directly to earlier discussions of work done against gravity and illustrates how even brief everyday motions can require substantial power output, comparable to a significant fraction of a horsepower.
The Verification
We verify the result by directly computing the work done against gravity and dividing by the time interval.
import numpy as np
# Given values
m = 0.9 * 80.0 # kg (effective lifted mass)
g = 9.81 # m/s^2
dy = 0.60 # m
t = 0.8 # s
# Work and power
W = m * g * dy
P = W / t
print("The average power supplied during the pull-up is %.0f W." % P)
7.4.2. Example Problem: Power Driving Uphill#
Exercise 7.11
The Problem
How much power must an automobile engine expend to move a \(1200\ \mathrm{kg}\) car up a \(15\%\) grade at \(90\ \mathrm{km/h}\)? Assume that \(25\%\) of this power is dissipated overcoming air resistance and friction.
Fig. 7.11 Image Credit: Openstax#
The Model
The car is treated as a point mass moving up an inclined plane at constant speed. Because the speed is constant, the car’s kinetic energy does not change, so the net work done on the car is zero. The engine’s power output is therefore balanced by the power expended against gravity and dissipative forces. The dissipative forces are modeled as a fixed fraction of the total power, with \(75\%\) of the engine’s power used to work against gravity and the remaining \(25\%\) lost to air resistance and friction.
The Math
At constant speed, the rate at which the engine does work against gravity equals the gravitational force component along the incline multiplied by the speed. The power required to lift the car at speed \(v\) up an incline of angle \(\theta\) is
A \(15\%\) grade means that the tangent of the incline angle is \(\tan\theta = 0.15\), so the angle is determined geometrically. Because only \(75\%\) of the engine’s power is used to overcome gravity, the total engine power \(P\) satisfies
We convert the speed to SI units and substitute the given values to evaluate the required power.
The Conclusion
We find that the engine must supply approximately \(5.8\times10^{4}\ \mathrm{W}\) of power to drive the car up the incline at constant speed. This power accounts not only for lifting the car against gravity but also for energy dissipated by air resistance and friction. The result connects directly to earlier discussions of work, power, and constant-speed motion, illustrating how engine power requirements depend on both gravitational forces and real-world energy losses in everyday driving situations.
The Verification
We verify the result by computing the gravitational power requirement and dividing by the fraction of power available to overcome gravity.
import numpy as np
# Given values
m = 1200.0 # kg
g = 9.81 # m/s^2
v = 25.0 # m/s (90 km/h)
theta = np.arctan(0.15)
fraction = 0.75 # fraction of power used against gravity
# Power calculation
P = m * g * v * np.sin(theta) / fraction
print("The required engine power is %.2e W." % P)
7.5. In-class Problems#
7.5.1. Part I#
Problem 1
A \(75\ \mathrm{kg}\) person climbs stairs, gaining \(2.50\ \mathrm{m}\) in height. Find the work done to accomplish this task.
Problem 2
A toy cart is pulled a distance of \(6.0\ \mathrm{m}\) in a straight line across the floor. The force pulling the cart has a magnitude of \(20\ \mathrm{N}\) and is directed at \(37^\circ\) above the horizontal. What is the work done by this force?
Problem 3
Calculate the kinetic energies of
(a) a \(2000\ \mathrm{kg}\) automobile moving at \(100\ \mathrm{km/h}\);
(b) an \(80\ \mathrm{kg}\) runner sprinting at \(10\ \mathrm{m/s}\); and
(c) a \(9.1 \times 10^{-31}\ \mathrm{kg}\) electron moving at \(2.0 \times 10^{7}\ \mathrm{m/s}\).
Problem 4
An \(8.0\ \mathrm{g}\) bullet has a speed of \(800\ \mathrm{m/s}\).
(a) What is its kinetic energy?
(b) What is its kinetic energy if the speed is halved?
7.5.2. Part II#
Problem 5
(a) Calculate the force needed to bring a \(950\)-kg car to rest from a speed of \(90\ \mathrm{km/h}\) in a distance of \(120\ \mathrm{m}\) (a fairly typical distance for a non-panic stop).
(b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in \(2.0\ \mathrm{m}\). Calculate the force exerted on the car and compare it with the force found in part (a).
Problem 6
A \(5.0\)-kg box has an acceleration of \(2.0\ \mathrm{m/s^2}\) when it is pulled by a horizontal force across a surface with \(\mu_K = 0.50\). Find the work done over a distance of \(10\ \mathrm{cm}\) by
(a) the horizontal force,
(b) the frictional force, and
(c) the net force.
(d) What is the change in kinetic energy of the box?
Problem 7
(a) How long will it take an \(850\)-kg car with a useful power output of \(40\ \mathrm{hp}\) (\(1\ \mathrm{hp} = 746\ \mathrm{W}\)) to reach a speed of \(15.0\ \mathrm{m/s}\), neglecting friction?
(b) How long will this acceleration take if the car also climbs a \(3.0\)-m high hill in the process?
Problem 8
An electron in a television tube is accelerated uniformly from rest to a speed of \(8.4 \times 10^{7}\ \mathrm{m/s}\) over a distance of \(2.5\ \mathrm{cm}\). What is the power delivered to the electron at the instant that its displacement is \(1.0\ \mathrm{cm}\)?
7.6. Homework#
7.6.1. Conceptual Problems#
Problem 1
One particle has mass \(m\) and a second particle has mass \(2m\). The second particle is moving with speed \(v\) and the first with speed \(2v\). How do their kinetic energies compare?
Problem 2
Two marbles of masses \(m\) and \(2m\) are dropped from a height \(h\). Compare their kinetic energies when they reach the ground.
7.6.2. Numerical Problems#
Problem 3
Calculate the work done by an \(85\ \mathrm{kg}\) man who pushes a crate \(4.0\ \mathrm{m}\) up along a ramp that makes an angle of \(20^\circ\) with the horizontal. He exerts a force of \(500\ \mathrm{N}\) on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate and on his body to get up the ramp.
Fig. 7.12 Image Credit: Openstax#
Problem 4
Compare the kinetic energy of a \(20{,}000\ \mathrm{kg}\) truck moving at \(110\ \mathrm{km/h}\) with that of an \(80\ \mathrm{kg}\) astronaut in orbit moving at \(27{,}500\ \mathrm{km/h}\).
Problem 5
A small block of mass \(200\ \mathrm{g}\) starts at rest at \(A\), slides to \(B\) where its speed is \(v_B = 8.0\ \mathrm{m/s}\), then slides along the horizontal surface a distance \(10\ \mathrm{m}\) before coming to rest at \(C\).
(a) What is the work of friction along the curved surface?
(b) What is the coefficient of kinetic friction along the horizontal surface?
Fig. 7.13 Image Credit: Openstax#
Problem 6
A \(500\ \mathrm{kg}\) dragster accelerates from rest to a final speed of \(110\ \mathrm{m/s}\) in \(400\ \mathrm{m}\) (about a quarter of a mile) and encounters an average frictional force of \(1200\ \mathrm{N}\). What is its average power output in watts and horsepower if this takes \(7.3\ \mathrm{s}\)?