 College Physics

# 16.1Hooke’s Law: Stress and Strain Revisited

College Physics16.1 Hooke’s Law: Stress and Strain Revisited
Figure 16.2 When displaced from its vertical equilibrium position, this plastic ruler oscillates back and forth because of the restoring force opposing displacement. When the ruler is on the left, there is a force to the right, and vice versa.

Newton’s first law implies that an object oscillating back and forth is experiencing forces. Without force, the object would move in a straight line at a constant speed rather than oscillate. Consider, for example, plucking a plastic ruler to the left as shown in Figure 16.2. The deformation of the ruler creates a force in the opposite direction, known as a restoring force. Once released, the restoring force causes the ruler to move back toward its stable equilibrium position, where the net force on it is zero. However, by the time the ruler gets there, it gains momentum and continues to move to the right, producing the opposite deformation. It is then forced to the left, back through equilibrium, and the process is repeated until dissipative forces dampen the motion. These forces remove mechanical energy from the system, gradually reducing the motion until the ruler comes to rest.

The simplest oscillations occur when the restoring force is directly proportional to displacement. When stress and strain were covered in Newton’s Third Law of Motion, the name was given to this relationship between force and displacement was Hooke’s law:

$F = − kx. F = − kx. size 12{F= - ital "kx"} {}$
16.1

Here, $FF size 12{F} {}$ is the restoring force, $xx size 12{x} {}$ is the displacement from equilibrium or deformation, and $kk size 12{k} {}$ is a constant related to the difficulty in deforming the system. The minus sign indicates the restoring force is in the direction opposite to the displacement.

Figure 16.3 (a) The plastic ruler has been released, and the restoring force is returning the ruler to its equilibrium position. (b) The net force is zero at the equilibrium position, but the ruler has momentum and continues to move to the right. (c) The restoring force is in the opposite direction. It stops the ruler and moves it back toward equilibrium again. (d) Now the ruler has momentum to the left. (e) In the absence of damping (caused by frictional forces), the ruler reaches its original position. From there, the motion will repeat itself.

The force constant $kk size 12{k} {}$ is related to the rigidity (or stiffness) of a system—the larger the force constant, the greater the restoring force, and the stiffer the system. The units of $kk size 12{k} {}$ are newtons per meter (N/m). For example, $kk size 12{k} {}$ is directly related to Young’s modulus when we stretch a string. Figure 16.4 shows a graph of the absolute value of the restoring force versus the displacement for a system that can be described by Hooke’s law—a simple spring in this case. The slope of the graph equals the force constant $kk size 12{k} {}$ in newtons per meter. A common physics laboratory exercise is to measure restoring forces created by springs, determine if they follow Hooke’s law, and calculate their force constants if they do.

Figure 16.4 (a) A graph of absolute value of the restoring force versus displacement is displayed. The fact that the graph is a straight line means that the system obeys Hooke’s law. The slope of the graph is the force constant $kk size 12{k} {}$. (b) The data in the graph were generated by measuring the displacement of a spring from equilibrium while supporting various weights. The restoring force equals the weight supported, if the mass is stationary.

### Example 16.1

#### How Stiff Are Car Springs?

Figure 16.5 The mass of a car increases due to the introduction of a passenger. This affects the displacement of the car on its suspension system. (credit: exfordy on Flickr)

What is the force constant for the suspension system of a car that settles 1.20 cm when an 80.0-kg person gets in?

#### Strategy

Consider the car to be in its equilibrium position $x=0x=0 size 12{x=0} {}$ before the person gets in. The car then settles down 1.20 cm, which means it is displaced to a position $x=−1.20×10−2mx=−1.20×10−2m size 12{x= - 1 "." "20" times "10" rSup { size 8{ - 2} } m} {}$. At that point, the springs supply a restoring force $FF size 12{F} {}$ equal to the person’s weight $w=mg=80.0 kg9.80m/s2=784Nw=mg=80.0 kg9.80m/s2=784N size 12{w= ital "mg"= left ("80" "." 0"kg" right ) left (9 "." "80""m/s" rSup { size 8{2} } right )="784"N} {}$. We take this force to be $FF size 12{F} {}$ in Hooke’s law. Knowing $FF size 12{F} {}$ and $xx size 12{x} {}$, we can then solve the force constant $kk size 12{k} {}$.

#### Solution

1. Solve Hooke’s law, $F=−kxF=−kx size 12{F= - ital "kx"} {}$, for $kk size 12{k} {}$:
$k = − F x . k = − F x . size 12{k= - { {F} over {x} } } {}$
16.2

Substitute known values and solve $kk size 12{k} {}$:

k = − 784 N − 1 . 20 × 10 − 2 m = 6 . 53 × 10 4 N/m. k = − 784 N − 1 . 20 × 10 − 2 m = 6 . 53 × 10 4 N/m. alignl { stack { size 12{k= - { {"784"" N"} over { - 1 "." "20" times "10" rSup { size 8{ - 2} } " m"} } } {} # =6 "." "53" times "10" rSup { size 8{4} } " N/m" {} } } {}
16.3

#### Discussion

Note that $FF size 12{F} {}$ and $xx size 12{x} {}$ have opposite signs because they are in opposite directions—the restoring force is up, and the displacement is down. Also, note that the car would oscillate up and down when the person got in if it were not for damping (due to frictional forces) provided by shock absorbers. Bouncing cars are a sure sign of bad shock absorbers.

### Energy in Hooke’s Law of Deformation

In order to produce a deformation, work must be done. That is, a force must be exerted through a distance, whether you pluck a guitar string or compress a car spring. If the only result is deformation, and no work goes into thermal, sound, or kinetic energy, then all the work is initially stored in the deformed object as some form of potential energy. The potential energy stored in a spring is $PEel=12kx2PEel=12kx2 size 12{"PE" rSub { size 8{"el"} } = { {1} over {2} } ital "kx" rSup { size 8{2} } } {}$. Here, we generalize the idea to elastic potential energy for a deformation of any system that can be described by Hooke’s law. Hence,

$PEel=12kx2,PEel=12kx2, size 12{"PE" size 8{"el"}= { {1} over {2} } ital "kx" rSup { size 8{2} } } {}$
16.4

where $PEelPEel size 12{"PE" rSub { size 8{"el"} } } {}$ is the elastic potential energy stored in any deformed system that obeys Hooke’s law and has a displacement $xx size 12{x} {}$ from equilibrium and a force constant $kk size 12{k} {}$.

It is possible to find the work done in deforming a system in order to find the energy stored. This work is performed by an applied force $FappFapp size 12{F rSub { size 8{"app"} } } {}$. The applied force is exactly opposite to the restoring force (action-reaction), and so $Fapp=kxFapp=kx size 12{F rSub { size 8{ ital "app"} } = ital "kx"} {}$. Figure 16.6 shows a graph of the applied force versus deformation $xx size 12{x} {}$ for a system that can be described by Hooke’s law. Work done on the system is force multiplied by distance, which equals the area under the curve or $(1/2)kx2(1/2)kx2 size 12{ $$1/2$$ ital "kx" rSup { size 8{2} } } {}$(Method A in the figure). Another way to determine the work is to note that the force increases linearly from 0 to $kxkx size 12{ ital "kx"} {}$, so that the average force is $(1/2)kx(1/2)kx size 12{ $$1/2$$ ital "kx"} {}$, the distance moved is $xx size 12{x} {}$, and thus $W=Fappd=[(1/2)kx](x)=(1/2)kx2W=Fappd=[(1/2)kx](x)=(1/2)kx2 size 12{W=F rSub { size 8{ ital "app"} } "." d= $$$1/2$$ ital "kx"$ $$x$$ = $$1/2$$ ital "kx" rSup { size 8{2} } } {}$ (Method B in the figure).

Figure 16.6 A graph of applied force versus distance for the deformation of a system that can be described by Hooke’s law is displayed. The work done on the system equals the area under the graph or the area of the triangle, which is half its base multiplied by its height, or $W=(1/2)kx2W=(1/2)kx2 size 12{W= $$1/2$$ ital "kx" rSup { size 8{2} } } {}$.

### Example 16.2

#### Calculating Stored Energy: A Tranquilizer Gun Spring

We can use a toy gun’s spring mechanism to ask and answer two simple questions: (a) How much energy is stored in the spring of a tranquilizer gun that has a force constant of 50.0 N/m and is compressed 0.150 m? (b) If you neglect friction and the mass of the spring, at what speed will a 2.00-g projectile be ejected from the gun?

Figure 16.7 (a) In this image of the gun, the spring is uncompressed before being cocked. (b) The spring has been compressed a distance $xx size 12{x} {}$, and the projectile is in place. (c) When released, the spring converts elastic potential energy $PEelPEel size 12{"PE" rSub { size 8{"el"} } } {}$ into kinetic energy.

#### Strategy for a

(a): The energy stored in the spring can be found directly from elastic potential energy equation, because $kk size 12{k} {}$ and $xx size 12{x} {}$ are given.

#### Solution for a

Entering the given values for $kk size 12{k} {}$ and $xx size 12{x} {}$ yields

PE el = 1 2 kx 2 = 1 2 50 . 0 N/m 0 . 150 m 2 = 0 . 563 N ⋅ m = 0 . 563 J PE el = 1 2 kx 2 = 1 2 50 . 0 N/m 0 . 150 m 2 = 0 . 563 N ⋅ m = 0 . 563 J alignl { stack { size 12{"PE" size 8{"el"}= { {1} over {2} } ital "kx" rSup { size 8{2} } = { {1} over {2} } left ("50" "." 0" N/m" right ) left (0 "." "150"" m" right ) rSup { size 8{2} } =0 "." "563"N cdot M} {} # =0 "." "563"J {} } } {}
16.5

#### Strategy for b

Because there is no friction, the potential energy is converted entirely into kinetic energy. The expression for kinetic energy can be solved for the projectile’s speed.

#### Solution for b

1. Identify known quantities:
$KE f = PE el or 1/2mv2=(1/2)kx2=PEel=0.563J KE f = PE el or KE_f = PE_el 1/2mv2=(1/2)kx2=PEel=0.563J size 12{1/2 ital "mv" rSup { size 8{2} } = $$1/2$$ ital "kx" rSup { size 8{2} } = ital "PE" rSub { size 8{e1} } =0 "." "563"J} {}$
16.6
2. Solve for $vv size 12{v} {}$:
$v = 2 PE el m 1 / 2 = 2 0 . 563 J 0 . 002 kg 1 / 2 = 23 . 7 J/kg 1 / 2 v = 2 PE el m 1 / 2 = 2 0 . 563 J 0 . 002 kg 1 / 2 = 23 . 7 J/kg 1 / 2 size 12{v= left [ { {2"PE" size 8{"el"}} over {m} } right ] rSup { size 8{1/2} } = left [ { {2 left (0 "." "563"" J" right )} over {0 "." "002"" kg"} } right ] rSup { size 8{1/2} } ="23" "." 7 left ("J/kg" right ) rSup { size 8{ { {1} over {2} } } } } {}$
16.7
3. Convert units: $23.7 m / s 23.7 m / s 23.7 m/s$

#### Discussion

(a) and (b): This projectile speed is impressive for a tranquilizer gun (more than 80 km/h). The numbers in this problem seem reasonable. The force needed to compress the spring is small enough for an adult to manage, and the energy imparted to the dart is small enough to limit the damage it might do. Yet, the speed of the dart is great enough for it to travel an acceptable distance.

Envision holding the end of a ruler with one hand and deforming it with the other. When you let go, you can see the oscillations of the ruler. In what way could you modify this simple experiment to increase the rigidity of the system?

If you apply a deforming force on an object and let it come to equilibrium, what happened to the work you did on the system?

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