#### Hooke's Law

__Extension and Compression__

__Extension and Compression__

- T = kx
- T is the magnitude of tension
- x is the extension or compression
- k is the spring constant
- Combining spring constants k:
- Springs in parallel:

- k
_{T}= k_{1}+ k_{2}+ ⋅⋅⋅

- k
- Springs in series:

^{1}/_{kT}=^{1}/_{k1}+^{1}/_{k2}+ ⋅⋅⋅

- Springs in parallel:

__Modulus of Elasticity__

__Modulus of Elasticity__

- T =
^{λx}/_{l} - λ is the modulus of elasticity
- l is the natural length
- Relating the spring constant and the modulus of elasticity: λ = kl

__Scenarios__

__Scenarios__

- If a mass is hanging at one end of a spring with the other attached to a fixed point, the tension in the spring must be equal to the weight of the object

- If two springs are attached between two fixed points and both are extended, the tension in both must be the same in order for there to be no overall net force

- If two springs are attached to fixed points and an object in the middle, tensions and frictional force must act in such a way that the overall force on the mass is equal to 0

- If a spring is attached to an object on a rough surface, the frictional force acts in the direction opposing tension in the spring (preventing it to return to its original shape)
- If a mass is on an incline and held at rest by a spring, the tension in the spring must be equal to the component of the weight parallel to the slope

__Elastic Potential Energy__

__Elastic Potential Energy__

- Work done in stretching (or compressing) a string or spring is given by W = ½kx²
- For a scenario, form an equation by conservation of energy and you can include elastic potential energy, kinetic energy, gravitational potential energy, and work done against friction

__Examples__

__Examples__

- A particle P of mass 0.35 kg is attached to the mid-point of a light elastic string of natural length 4 m. Ends of the string are attached to fixed points A and B. P hangs in equilibrium 0.7 m vertically below the mid-point M of AB

- Find the tension in the string and hence show that the modulus of elasticity of the string is 25 N
- Find the extension of the string by finding length AP and PB using right angled triangles:
- AP = PB = √(2.4² + 0.7²) = 2.5 m
- 2.5 + 2.5 – 4 = 1 m

- The vertical component of tension in the string is equal to the weight as the system is in equilibrium:
- 2T cos θ = mg
- 2T × (
^{0.7}/_{2.5}) = 0.35 × 10 - ∴ T = 6.25

- Find the modulus of elasticity by using the calculated information:
- T =
^{λx}/_{l} - λ = 6.25 × 4 = 25

- T =

- Find the extension of the string by finding length AP and PB using right angled triangles:
- P is now held at 1.8 m vertically below M, and released. Find the speed with which P passes through M
- Find the spring constant k:
- k =
^{λ}/_{l}=^{25}/_{4}= 6.25

- k =
- Find the extension of the string
- AP = PB = √(2.4² + 1.8²) = 3 m
- 3 + 3 – 4 = 2 m

- Form an equation by the conservation of energy:
- e.p.e. = gain in k.e. + gain in g.p.e + e.p.e at M
- (½ × 6.25 × 2²) = (½ × 0.35 × v²) + (0.35 × 10 × 1.8) + (½ × 6.25 × 0.8²)
- 12.5 = 0.175v² + 6.3 + 2
- v² = 24
- ∴ v = ±4.90 ms
^{-1}

- Find the spring constant k:

- Find the tension in the string and hence show that the modulus of elasticity of the string is 25 N