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
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
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