HOOKES LAW

Physics

HOOKE’S LAW

Hooke’s law states that “the extension of a spring is proportional to the applied force, provided that the force is not large enough to deform the spring permanently”.

Mathematically expressed as Force α extension.

Spring constant

Since Force α extension then Force / Extension = constant (k). The constant of proportionality (k) is called the spring constant. F / e = k or Force (N) = k e. The spring constant is a measure of the stiffness of a spring.

The greater the constant the stiffer the spring.

The spring constant varies with the following;-

  1. a) Material – identical springs mad of different materials will have different constants i.e.steel and copper.
  2. b) Diameter – the stiffness decreases with the increase in diameter.
  3. c) Thickness of the wire – a spring made of a thicker wire is stiffer than the one made of thin wire of the same material.
  4. d) Length of spring – a short spring is stiffer than a longer one.
  5. e) Number of turns per unit length – a spring with higher number of turns per unit length is less stiff than the one with fewer turns per unit length.
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Example

  1. If the springs shown below are similar and the constant of proportionality (k) is 100 Nm-1, determine total extension in each arrangement.

 

The spring balance

It is made up of a spring mounted in a metal or plastic casing.

The spring is fitted with a pointer which moves along a calibrated scale divided into ten equal parts.

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Examples

  1. A load of 4 N causes a certain copper wire to extend by 1.0 mm. Find the load that will cause a 3.2 mm extension on the same wire. (Assume Hooke’s law is obeyed).

Solution

F α e also F1 / F2 = e1 / e2 = F2 = (4 × 3.2) / 1.0 = 12.8 N. 2. A body of 200 g was hung from the lower end of a spring which obeys Hooke’s law. Given that the spring extended by 100 mm, what is the spring constant for this spring? Solution F = α e, F = k e. F = 200 × 10-3 kg × 10 N /kg = 2 N. Extension = 100 × 103 m = 0.1 m.

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Spring constant (k) = 2 / 0.1 = 20 N/m.

  1. Two identical springs, whose spring constant is 6.0 N/cm, are used to support a load of 60 N as shown below. Determine the extension of each spring.

 

Solution

Since the springs are parallel their spring constant equals 2k. Therefore extension = Force / k = 2 F / k = 60 / 2 × 6 = 5 cm. Each spring will extend by 5 cm.

 

See also:

MAGNETIC EFFECT OF AN ELECTRIC CURRENT

REFLECTION AT CURVED SURFACES

EQUILIBRIUM AND CENTRE OF GRAVITY

TURNING EFFECT OF A FORCE

MEASUREMENT II

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