All flashcards topics

Elastic energy and the work-energy principleEdexcel A-Level Further Maths: Flashcards

Card 1 of 120 of 12 known

Question

State Hooke's law for an elastic string.

Tap or press Space to reveal

Tap card or press Space to flip

See all 12 cards
State Hooke's law for an elastic string.
T=λxlT=\frac{\lambda x}{l}, where xx is the extension and ll the natural length.
What does the modulus of elasticity λ\lambda measure, and in what units?
The stiffness of the string for its natural length; newtons.
State the formula for elastic potential energy.
λx22l=12Tx\frac{\lambda x^2}{2l}=\frac12Tx
Why is elastic energy 12Tx\frac12Tx and not TxTx?
The tension rises from 0 to TT, so the work done is the area of a triangle, 12Tx\frac12Tx.
How much energy is stored when a string is slack?
None; the tension is zero.
Work done stretching a string from extension x1x_1 to x2x_2?
λ2l(x22−x12)\frac{\lambda}{2l}\left(x_2^2-x_1^2\right)
State the work-energy principle for elastic problems.
Work done by other forces (e.g. friction) = change in (KE + GPE + EPE).
When is mechanical energy conserved?
When only gravity and the elastic force do work (no friction or other external forces).
Work done against friction?
F×F\times distance moved, with F=μRF=\mu R.
A vertical string is released from rest at its fixing point. How far has the particle fallen at greatest extension xx?
l+xl+x, the natural length plus the extension.
How does a spring differ from a string?
A spring can be compressed and then pushes; a string goes slack.
A particle on a smooth table is released from extension xx. What is its speed when the string goes slack?
Solve λx22l=12mv2\frac{\lambda x^2}{2l}=\frac12mv^2.

Exam questions on Elastic energy and the work-energy principle

  1. A light elastic string has natural length 0.80.8 m and modulus of elasticity 4040 N. It is stretched so that its extension is 0.20.2 m.
    The string is stretched further until its extension is 0.40.4 m. Calculate the additional work done in stretching it.2 marks
  2. A particle 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 147147 N. The other end of the string is fixed to a point OO. Take g=9.8g=9.8 m s−2^{-2}.
    The particle is now held at rest with the string just taut, and then released. Find the speed of the particle when it passes through the equilibrium position.2 marks
  3. A particle PP of mass 0.40.4 kg lies on a horizontal table. It is attached to one end of a light elastic string of natural length 0.60.6 m and modulus of elasticity 2424 N. The other end of the string is fixed to a point AA on the table. PP is pulled to a point 11 m from AA, with the string stretched, and released from rest. Take g=9.8g=9.8 m s−2^{-2}.
    The table is smooth. Find the speed of PP when the string becomes slack.3 marks
See the full worksheet

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).