 4 Ijesha Close, Ilupeju, Lagos
+2347 086 296 002

#### Hooke's Law

###### 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: • kT = k1 + k2 + ⋅⋅⋅
• Springs in series: • 1/kT = 1/k1 + 1/k2 + ⋅⋅⋅
###### 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
• 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
• 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
1. 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 1. 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
2. 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
• 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
Minimum 4 characters