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Elastic potential energyEdexcel International A Level Further Maths: Flashcards

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Elastic potential energy stored in a string or spring?

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Elastic potential energy stored in a string or spring?
λx22l\frac{\lambda x^2}{2l}, where xx is the extension
Tension in an elastic string?
T=λxlT=\frac{\lambda x}{l}
Elastic energy in terms of tension?
12Tx\frac12Tx
Why is the work done in stretching λx22l\frac{\lambda x^2}{2l} and not λx2l\frac{\lambda x^2}{l}?
The tension increases from 00 to λxl\frac{\lambda x}{l}, so the work is the area under a graph: ∫0xλslds\int_0^x\frac{\lambda s}{l}ds.
Unit of the modulus of elasticity λ\lambda?
Newtons (N)
Elastic energy when a string is slack?
Zero
Work done changing extension from x1x_1 to x2x_2?
λ2l(x22−x12)\frac{\lambda}{2l}(x_2^2-x_1^2)
Energy stored when a spring is compressed by xx?
λx22l\frac{\lambda x^2}{2l}, the same as for extension xx
Work-energy principle (with elastic energy)?
Work done by other forces = change in (KE + GPE + EPE)
When is mechanical energy conserved?
When there is no friction, resistance or other external work.
At the greatest extension of a string, what is the speed?
Zero, so the energy is all elastic and gravitational.
Where is the speed maximum for a particle on a string?
Where the resultant force is zero (acceleration zero).
Loss of GPE for a particle moving a distance dd down a smooth slope at angle α\alpha?
mgdsin⁡αmgd\sin\alpha

Exam questions on Elastic potential energy

  1. A light elastic string has natural length 0.60.6 m and modulus of elasticity 3030 N. The string is stretched to a length of 0.90.9 m.
    The string is now stretched slowly from a length of 0.90.9 m to a length of 1.11.1 m. Find the work done against the tension of the string.2 marks
  2. A particle PP of mass 0.50.5 kg is attached to one end of a light elastic spring of natural length 0.40.4 m and modulus of elasticity 2020 N. The other end of the spring is fixed to a point OO on a smooth horizontal table. PP is held on the table at a distance 0.60.6 m from OO and released from rest.
    Find the speed of PP when it is 0.50.5 m from OO.2 marks
  3. A particle PP of mass 22 kg is attached to one end of a light elastic string of natural length 1.51.5 m and modulus of elasticity 6060 N. The other end of the string is fixed to a point AA on a ceiling. PP is released from rest at AA and falls vertically. Take g=9.8g=9.8 m s−2^{-2} and ignore air resistance.
    Show that, when PP first comes to instantaneous rest, the extension xx metres of the string satisfies 20x2−19.6x−29.4=020x^2-19.6x-29.4=0.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).