Physics Form 2 Notes : CHAPTER SEVEN – HOOKE’S LAW.

CHAPTER SEVEN

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

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.

Solution

1. k = 100 Nm^-1, extension = force / k = 10 / 100 = 0.1 m = 10 cm.
2. Extension of the lower spring = 10 cm, extension of the two parallel springs = 5 cm. Total extension = 10 + 5 = 15 cm.
3. Extension of the two lower springs = 5 cm. Middle spring extends by 10 cm. Upper two springs extend by 5 cm. Total extension = 5 + 10 + 5 = 20 cm.

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.

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

1. Since F α e, then F₁ / F₂ = e₁ / e₂. Therefore, 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

1. F = k e. F = 200 × 10^-3 kg × 10 N/kg = 2 N.

Extension = 100 × 10^-3 m = 0.1 m.

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

3. 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 combined spring constant equals 2k.

Therefore, extension = Force / (2k) = 60 / (2 × 6) = 5 cm. Each spring will extend by 5 cm.