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The graph here represents applied force, given along y-axis, versus deformation or displacement, given along x axis. The slope is linear slanting and the slope area is covered between x axis and the slope, given by F is equal to k multiplied by x, where k is constant and x is displacement. The force applied along y-axis is given by half of k multiplied by x. Along with the graph, two methods are provided to calculate weight, W. The first method gives the solution by multiplying half of b multiplied by h, whereas in the second we can get the solution by multiplying f with x.
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 ) kx 2 size 12{W= \( 1/2 \) ital "kx" rSup { size 8{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?

The figure a shows an artistic impression of a tranquilizer gun, which shows the inside of it revealing the gun spring and a panel just below it, in the outside area, attached to the spring. This stage shows the gun before it is cocked, and the spring is uncompressed covering the entire inside area. The figure b shows the gun with the spring in the compressed mode. The spring has been compressed to a distance x, where x distance shows the vacant area inside the gun through which the spring has been compressed. The panel is also moving along the spring. And a bullet of mass m is shown at the front of the compressed spring. The spring here has elastic potential energy, represented by P E sub e l. The figure c is the third stage of the above two stages of the gun. The spring here is released from the compressed stage releasing the bullet in the outer forward direction with velocity V and the spring’s potential energy is converted into kinetic energy, represented here by K E.
(a) In this image of the gun, the spring is uncompressed before being cocked. (b) The spring has been compressed a distance x size 12{x} {} , and the projectile is in place. (c) When released, the spring converts elastic potential energy PE el 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 k size 12{k} {} and x size 12{x} {} are given.

Solution for a

Entering the given values for k size 12{k} {} and x 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 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 {} } } {}

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 KE_f = PE_el 1 / 2 mv 2 = ( 1 / 2 ) kx 2 = PE el = 0 . 563 J 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} {}
  2. Solve for v 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 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} } } } } {}
  3. Convert units: 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.

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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?

Answer

You could hold the ruler at its midpoint so that the part of the ruler that oscillates is half as long as in the original experiment.

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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?

Answer

It was stored in the object as potential energy.

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Section summary

  • An oscillation is a back and forth motion of an object between two points of deformation.
  • An oscillation may create a wave, which is a disturbance that propagates from where it was created.
  • The simplest type of oscillations and waves are related to systems that can be described by Hooke’s law:
    F = kx , size 12{F= - ital "kx"} {}

    where F size 12{F} {} is the restoring force, x size 12{x} {} is the displacement from equilibrium or deformation, and k size 12{k} {} is the force constant of the system.

  • Elastic potential energy PE el size 12{"PE" rSub { size 8{"el"} } } {} stored in the deformation of a system that can be described by Hooke’s law is given by
    PE el = ( 1 / 2 ) kx 2 size 12{ ital "PE" rSub { size 8{e1} } = \( 1/2 \) ital "kx" rSup { size 8{2} } } {} .

Conceptual questions

Describe a system in which elastic potential energy is stored.

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Problems&Exercises

Fish are hung on a spring scale to determine their mass (most fishermen feel no obligation to truthfully report the mass).

(a) What is the force constant of the spring in such a scale if it the spring stretches 8.00 cm for a 10.0 kg load?

(b) What is the mass of a fish that stretches the spring 5.50 cm?

(c) How far apart are the half-kilogram marks on the scale?

(a) 1 . 23 × 10 3 N/m size 12{1 "." "23" times "10" rSup { size 8{3} } `"N/m"} {}

(b) 6 . 88 kg size 12{6 "." "88"`"kg"} {}

(c) 4 .00 mm size 12{4 "." "00"`"mm"} {}

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It is weigh-in time for the local under-85-kg rugby team. The bathroom scale used to assess eligibility can be described by Hooke’s law and is depressed 0.75 cm by its maximum load of 120 kg. (a) What is the spring’s effective spring constant? (b) A player stands on the scales and depresses it by 0.48 cm. Is he eligible to play on this under-85 kg team?

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One type of BB gun uses a spring-driven plunger to blow the BB from its barrel. (a) Calculate the force constant of its plunger’s spring if you must compress it 0.150 m to drive the 0.0500-kg plunger to a top speed of 20.0 m/s. (b) What force must be exerted to compress the spring?

(a) 889 N/m

(b) 133 N

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(a) The springs of a pickup truck act like a single spring with a force constant of 1 . 30 × 10 5 N/m size 12{1 "." "30" times "10" rSup { size 8{5} } } {} . By how much will the truck be depressed by its maximum load of 1000 kg?

(b) If the pickup truck has four identical springs, what is the force constant of each?

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When an 80.0-kg man stands on a pogo stick, the spring is compressed 0.120 m.

(a) What is the force constant of the spring? (b) Will the spring be compressed more when he hops down the road?

(a) 6 . 53 × 10 3 N/m size 12{ {underline {6 "." "53" times "10" rSup { size 8{5} } " N/m"}} } {}

(b) Yes

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A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it. (a) What is the force constant of the spring? (b) What is the unloaded length of the spring?

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Practice Key Terms 4

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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