Further Mechanics: Work, Energy and Elastic Strings Depth
Work, Energy and Elastic Strings Depth, a Further Mechanics topic, combines the work-energy principle and elastic strings or springs into richer problems where kinetic energy, gravitational potential energy, elastic potential energy and, where present, work done against friction or resistance must all be tracked together across a single motion. It also introduces finding the point of GREATEST SPEED during such a motion: unlike the point of maximum extension, which is found from an energy argument (speed momentarily zero), the point of greatest speed is found from a force argument (the resultant force, and so the acceleration, is momentarily zero there), typically where the tension or thrust exactly balances the other forces resolved along the direction of motion.
Before you start
Make sure you're comfortable with these topics first:
Method
- List every force doing work on the particle: gravity, any elastic tension or thrust, and friction or resistance if present. Gravity and the elastic force are conservative, so can be tracked with GPE and EPE terms; friction and resistance are not, and must be tracked as work done directly.
- Write the full energy equation between the two points of interest: (KE+GPE+EPE) at the start = (KE+GPE+EPE) at the end + work done against friction or resistance between the two points.
- Take care with the sign of the GPE term, using +mgh for a rise and -mgh for a fall relative to a chosen zero level, applied consistently.
- Take care with the EPE term: it is only non-zero while a string is taut (extension x>0) or a spring is stretched or compressed; treat any phase where a string is slack as ordinary motion under gravity (and friction or resistance, if present) alone, with EPE=0 there.
- To find the point of greatest speed, set the resultant force along the direction of motion to zero: for a particle hanging on a string or spring under gravity, this is where the tension (or thrust) exactly balances the weight (or the weight's component along the line of motion, plus or minus any constant friction or resistance).
- Use the position found this way, together with the energy equation from the starting point, to find the greatest speed itself, without needing calculus.
- Where the motion has more than one phase (for example free motion while a string is slack, followed by elastic motion once it becomes taut), apply the energy equation across the whole motion in one go if every force's work over the full distance can be found; a resistance or friction force that acts throughout, regardless of the string's state, does work over the full distance travelled, while the EPE term applies only to the taut phase.
Worked example
A particle P of mass 0.5 kg hangs from a fixed point O by a light elastic string of natural length 1.6 m and modulus of elasticity 19.6 N. P is held at the point where the string is exactly at its natural length and released from rest, so that it moves vertically downward. As P falls, find the greatest speed it reaches, and the extension of the string at the instant this greatest speed occurs. (Take g = 9.8 m/s^2.)
- The particle's speed is greatest where its acceleration is zero, i.e. where the resultant force on it is zero: this is where the tension in the string equals its weight, T=mg.
- Apply Hooke's law at this point: lambda*x/L=mg, so 19.6*x/1.6=0.5(9.8)=4.9, giving 12.25x=4.9, so x=0.4 m.
- Use conservation of energy from the start (x=0, v=0, at the natural length) to this point (x=0.4, v=v_max): loss in GPE = gain in KE + gain in EPE, i.e. mg*x=(1/2)m*v_max^2+lambda*x^2/(2L).
- Substitute the known values: 0.5(9.8)(0.4) = (1/2)(0.5)v_max^2 + 19.6(0.4)^2/(2x1.6), i.e. 1.96 = 0.25v_max^2 + 0.98.
- Solve: 0.25v_max^2 = 1.96-0.98 = 0.98, so v_max^2 = 3.92, giving v_max = sqrt(3.92) = 7sqrt(2)/5 m/s, approximately 1.98 m/s (3 s.f.), with the string extended 0.4 m at this instant.
Practice questions
Try each question, then tap to reveal the answer.
Q1State the condition, in terms of forces, for the speed of a particle moving vertically on an elastic string to be at a maximum.Show answer
Answer: The resultant force on the particle is zero at that instant (zero acceleration), which for a particle hanging on a string under gravity alone means the tension equals the weight, T=mg.
Q2A particle of mass 0.2 kg hangs from an elastic string of natural length 0.5 m and modulus 9.8 N. Find the extension of the string at the point where the particle's speed, during a vertical motion, is greatest.Show answer
Answer: T=mg: 9.8x/0.5 = 0.2(9.8) = 1.96, so 19.6x=1.96, giving x=0.1 m.
Q3State, in words, the full energy equation used for a particle that moves under gravity, connected to an elastic string, on a rough surface.Show answer
Answer: (KE+GPE+EPE) at the start equals (KE+GPE+EPE) at the end, plus the work done against friction between the two points, since friction removes mechanical energy from the system rather than storing it.
Q4A block of mass 3 kg on a smooth horizontal surface is attached to a spring of natural length 0.6 m and modulus 30 N, fixed to a wall. It is pulled so the spring extends to 0.9 m, then released from rest. Find its speed as it passes through the point where the spring returns to its natural length.Show answer
Answer: Extension released from x=0.3 m gives EPE=30(0.3)^2/(2x0.6)=2.25 J, which converts entirely to KE on the smooth horizontal surface: (1/2)(3)v^2=2.25, so v^2=1.5 and v=sqrt(1.5)=sqrt(6)/2, approximately 1.22 m/s.
Q5Explain why, for a particle released from rest at the natural length of a vertical elastic string, the point of maximum speed is not the same as the point of maximum extension.Show answer
Answer: Maximum extension occurs where the speed is momentarily zero (all the energy has returned to GPE and EPE), whereas maximum speed occurs where the resultant force, and so the acceleration, is momentarily zero; these are different conditions (v=0 as against a=0), so they occur at different extensions.
Q6A particle of mass 0.6 kg hangs in equilibrium from an elastic string of natural length 1 m and modulus 14.7 N. Find the equilibrium extension, and state how it compares with the extension at the point of maximum speed if the particle were instead released from the natural length and allowed to fall.Show answer
Answer: At equilibrium, T=mg: 14.7x/1=0.6(9.8)=5.88, so x=0.4 m. This is exactly the same extension as the point of maximum speed found if the particle is released from the natural length, since both conditions require T=mg.
Q7A block of mass 2 kg on a rough horizontal surface (coefficient of friction 0.5) is pushed against a spring of natural length 0.5 m and modulus of elasticity 74 N, compressing it by 0.2 m, then released from rest. Find the speed of the block at the instant the spring returns to its natural length. (Take g = 9.8 m/s^2.)Show answer
Answer: EPE released = 74(0.2)^2/(2x0.5) = 2.96 J; work done against friction over 0.2 m = (0.5)(2)(9.8)(0.2) = 1.96 J; remaining KE = 2.96-1.96 = 1.00 J = (1/2)(2)v^2, so v^2=1 and v=1 m/s.
Q8The block in the previous question separates from the spring at its natural length and continues under friction alone. Find the additional distance it travels before coming to rest.Show answer
Answer: Deceleration = friction force/mass = 9.8/2 = 4.9 m/s^2. Using v^2=u^2-2as with u=1, v=0: 0=1-2(4.9)s, so s=1/9.8=5/49, approximately 0.102 m (3 s.f.).
Exam-style questions
Written in the style of a A Level Further Maths exam paper, with a full mark scheme.
One end of a light elastic string of natural length 1.5 m and modulus of elasticity 9.8 N is attached to a fixed point O. A particle of mass 0.6 kg is attached to the other end and is held with the string just taut, then released from rest and falls vertically. (a) Find the extension of the string at the point where the particle's speed is greatest, and find this greatest speed in exact surd form. (b) Find the maximum extension of the string during the motion, and verify that it is exactly twice the extension found in part (a). (Take g = 9.8 m/s^2.)
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A particle of mass 0.5 kg is attached to one end of a light elastic string of natural length 1 m and modulus of elasticity 14 N; the other end is fixed to a point O. The particle is held at O and released from rest, falling vertically through a resistant medium that exerts a constant resistance of 2 N to its motion whenever it moves. (a) Find the speed of the particle at the instant it has fallen a total distance of 1.4 m (that is, when the string's extension is 0.4 m). (b) Explain why the resistance does work over the whole 1.4 m fall, even though the string is only taut for part of it. (Take g = 9.8 m/s^2.)
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