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The given figure a is the graph of restoring force versus displacement. The displacement is given by x in meters along x axis, with scales from zero to point zero five zero, then to point one zero, then forward. The restoring force is given by F in unit newton along y axis, with scales from zero to two point zero to four point zero to forward. The graph line starts from zero and goes to upward to point where x is greater than point one zero and F is greater than four point zero with intersection dots at equal distances on the slope line. The slope is depicted by K which is given by rise along y-axis upon run along x axis . The values of mass in kilogram, weight in newtons, and displacement in meters are given along with the graph in a tabular format. In the figure b a horizontal weight bar is shown with three weight measuring springs tied to its lower part, hanging in the downward vertical direction. The first bar has no mass hanging through it, showing zero displacement, as x is equal to zero. It is the least stretched spring downward. The second spring has mass m one tied to it which exerts a force w one, on the spring, which causes displacement in the spring shown here to be x one. Similarly, the third spring is most stretched downward with a mass m two hanging through it with force w two and displacement x two. The values of mass in kg, weight in newtons and displacement in meters are given with the graph in a tabular format.
(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 k 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.

How stiff are car springs?

The figure shows the left side of a hatchback car’s back area, showing the font of its rear wheel. There is an arrow on road pointing its head toward this wheel.
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 = 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 2 m size 12{x= - 1 "." "20" times "10" rSup { size 8{ - 2} } m} {} . At that point, the springs supply a restoring force F size 12{F} {} equal to the person’s weight w = mg = 80 . 0 kg 9 . 80 m/s 2 = 784 N 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 F size 12{F} {} in Hooke’s law. Knowing F size 12{F} {} and x size 12{x} {} , we can then solve the force constant k size 12{k} {} .

Solution

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

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

    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" {} } } {}

Discussion

Note that F size 12{F} {} and x 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 PE el = 1 2 kx 2 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,

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

where PE el 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 x size 12{x} {} from equilibrium and a force constant k 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 F app size 12{F rSub { size 8{"app"} } } {} . The applied force is exactly opposite to the restoring force (action-reaction), and so F app = kx size 12{F rSub { size 8{ ital "app"} } = ital "kx"} {} . [link] shows a graph of the applied force versus deformation x 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 ) kx 2 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 kx size 12{ ital "kx"} {} , so that the average force is ( 1 / 2 ) kx size 12{ \( 1/2 \) ital "kx"} {} , the distance moved is x size 12{x} {} , and thus W = F app d = [ ( 1 / 2 ) kx ] ( x ) = ( 1 / 2 ) kx 2 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).

Practice Key Terms 4

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Source:  OpenStax, Yupparaj english program physics corresponding to thai physics book #3. OpenStax CNX. May 19, 2014 Download for free at http://legacy.cnx.org/content/col11657/1.1
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