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A-Level · Further Mechanics (Work, Energy, Elastic Strings/Springs and Circular Motion Depth)

FP.FM6 Further Mechanics: Work, Energy and Elastic Strings Depth

EDEXCEL 9FM0 · Calculator allowed · about 135 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Key results: work-energy principle, elastic strings/springs and circular motion

Original content written for Revision Library.

Unless a question states otherwise, take g = 9.8 m/s^2, model each object as a particle, and treat every string or spring as light; strings are inextensible while rods are light and rigid. This worksheet is a depth paper combining three areas of Further Mechanics, momentum, impulse and collisions are assumed known from elsewhere and are not the focus here: the work-energy principle, elastic strings and springs, and circular motion, and several questions require you to combine two of these ideas within a single question. Kinetic energy is KE = (1/2)*m*v^2 and gravitational potential energy relative to a reference level is GPE = m*g*h; the work-energy principle states that the total work done by all the forces acting on a particle (including work done against resistances such as friction) equals its change in kinetic energy. Hooke's law states that the tension (or thrust) in an elastic string or spring of natural length l and modulus of elasticity lambda, when extended (or compressed) by x, has magnitude T = lambda*x/l, and the elastic potential energy stored is EPE = lambda*x^2/(2*l); an elastic string can only pull, and goes slack once its length is less than l, whereas a spring can both pull and push. For a particle moving on a circle of radius r with angular speed omega, the linear speed is v = r*omega and the centripetal acceleration, directed towards the centre, has magnitude a = v^2/r = r*omega^2; for a conical pendulum, or a vehicle on a banked track, resolve forces vertically and horizontally to link the tension or normal reaction to the speed. For vertical circular motion, combine Newton's second law in the radial direction with energy conservation; for a particle on a string, or on the inside of a smooth track, the critical condition is that the tension or normal reaction is zero, whereas a particle threaded on a rigid rod can maintain circular motion even when this condition would otherwise be violated, in which case the rod is in thrust (compression) rather than tension.

A light elastic string has natural length 1.6 m. One end of the string is attached to a fixed point A on a ceiling. A particle of mass 3 kg is attached to the other end of the string and hangs in equilibrium vertically below A, with the string stretched to a length of 2.0 m. The string obeys Hooke's law with modulus of elasticity λ newtons. Take g = 9.8 m/s² throughout.

Figure (to be drawn): Diagram: fixed point A at the top, elastic string AP hanging vertically, extended to a total length of 2.0 m, particle P attached at the lower end with weight mg acting vertically downward.

(a)Find the extension of the string in this position.(1)

(b)By modelling P as being in equilibrium, use Hooke's law to find the value of λ.(3)

(Total for Question 1 is 4 marks)

A light elastic spring has natural length 0.5 m and modulus of elasticity 40 N. The spring is compressed, without buckling, to a length of 0.35 m.

Figure (to be drawn): Diagram: spring of natural length 0.5 m shown compressed to a length of 0.35 m, with the compression x marked between the free end and the natural-length position.

(a)State the compression of the spring.(1)

(b)Calculate the elastic potential energy (EPE) stored in the spring.(3)

(c)State one modelling assumption used in part (b) about the spring.(1)

(Total for Question 2 is 5 marks)

A block of mass 5 kg is pulled from rest along a rough horizontal floor by a constant horizontal force of 30 N. The coefficient of friction between the block and the floor is 0.25. The block travels a distance of 4 m. Take g = 9.8 m/s^2.

(a)Find the work done against friction as the block travels the 4 m.(2)

(b)Use the work-energy principle to find the speed of the block after it has travelled the 4 m, giving your answer to 3 significant figures.(4)

(Total for Question 3 is 6 marks)

A particle P of mass 0.4 kg is attached to one end of a light elastic string of natural length 0.6 m and modulus of elasticity 24 N. The other end of the string is attached to a fixed point O on a smooth horizontal table. P is held on the table at the point where OP is a straight line and the string is stretched to a length of 0.9 m, and is then released from rest.

(a)Find the extension of the string in P's initial position.(1)

(b)Find the elastic potential energy (EPE) stored in the string initially.(3)

(c)Given that the table is smooth, find the speed of P at the instant the string first becomes slack, ft their answer to (b).(3)

(Total for Question 4 is 7 marks)

A particle of mass 0.5 kg is attached to one end of a light inextensible string of length 0.75 m. The other end of the string is attached to a fixed point O. The particle moves in a complete vertical circle of radius 0.75 m about O, with the string remaining taut throughout the motion. Take g = 9.8 m/s^2.

Figure (to be drawn): Diagram: vertical circle of radius r = 0.75 m, centre O, with the particle shown at the top (tension T and weight mg both acting vertically downward, towards O) and at the bottom of the circle.

(a)By considering the particle at the top of the circle, show that the minimum possible speed there is given by v² = gr, and hence find this minimum speed, giving your answer to 3 significant figures.(4)

(b)Given that the particle has this minimum speed at the top of the circle, use the work-energy principle to find its speed at the bottom of the circle, giving your answer to 3 significant figures.(4)

(Total for Question 5 is 8 marks)

A car of mass 900 kg, driven by Priya, travels at a constant speed round a circular bend of radius 50 m on a country road near Matlock in Derbyshire. The road is banked at an angle θ to the horizontal, where tan(θ) = 0.3. The road is modelled as smooth, and the car is modelled as a particle travelling at exactly the speed for which no sideways friction force is needed to keep it on the bend. Take g = 9.8 m/s^2.

Figure (to be drawn): Diagram: cross-section of the banked road showing the angle θ to the horizontal, the normal reaction N acting perpendicular to the road surface, and the weight mg acting vertically downward; the centre of the circular bend lies to one side, in the direction of the horizontal component of N.

(a)Find the value of θ, giving your answer in degrees to 3 significant figures.(2)

(b)By resolving the forces on the car vertically and horizontally, show that v² = r*g*tan(θ), where v is the car's speed and r is the radius of the bend.(3)

(c)Hence find the speed, in m/s, at which Priya should drive round the bend.(3)

(Total for Question 6 is 8 marks)

A particle P of mass 0.3 kg is attached to one end of a light inextensible string of length 0.5 m. The other end of the string is attached to a fixed point O. P moves with constant angular speed ω in a horizontal circle, with the string at a constant angle of 40 degrees to the downward vertical through O (a conical pendulum). Take g = 9.8 m/s^2.

Figure (to be drawn): Diagram: fixed point O at the top, string OP at 40 degrees to the downward vertical, P moving in a horizontal circle of radius r directly below O, tension T shown acting along PO and weight mg acting vertically downward on P.

(a)By resolving vertically, show that the tension in the string is given by T = mg/cos(40 degrees), and find the value of T, giving your answer to 3 significant figures.(3)

(b)Find the radius r of the circle in which P moves, giving your answer to 3 significant figures.(2)

(c)By considering the horizontal equation of motion of P, find the value of ω, and hence find the speed of P, giving each answer to 3 significant figures.(4)

(Total for Question 7 is 9 marks)

A particle P of mass 1.5 kg lies on a rough plane inclined at 30 degrees to the horizontal. The coefficient of friction between P and the plane is 0.2. A light spring of natural length 0.9 m and modulus of elasticity 90 N lies along a line of greatest slope of the plane, with its lower end fixed to a point A at the bottom of the plane. P is attached to the upper end of the spring and is held at rest at the point where the spring is compressed to a length of 0.6 m. P is then released from rest and moves up the plane. Take g = 9.8 m/s^2.

Figure (to be drawn): Diagram: plane inclined at 30 degrees to the horizontal, spring AP lying along a line of greatest slope with A fixed at the bottom, P shown at its initial compressed position; the weight component mg*sin(30) down the slope, the normal reaction, and the friction force (acting down the slope as P moves up) are all marked.

(a)Find the elastic potential energy (EPE) stored in the spring in P's initial position.(3)

(b)Find the work done against friction, and the work done against gravity, as P moves from its initial position to the point where the spring first reaches its natural length (a distance of 0.3 m up the slope).(4)

(c)Use the work-energy principle to find the speed of P at the instant the spring reaches its natural length, giving your answer to 3 significant figures, ft your answers to (a) and (b).(3)

(Total for Question 8 is 10 marks)

At an engineering workshop in Coventry, a technician called Dele is testing a model rotating-arm ride. A particle P of mass 0.25 kg is attached to one end of a light rigid rod OP of length 0.4 m; the other end of the rod is freely hinged at a fixed point O, and P moves in a complete vertical circle of radius 0.4 m about O. At the lowest point of the circle, P has speed 4 m/s. Take g = 9.8 m/s^2.

Figure (to be drawn): Diagram: vertical circle of radius r = 0.4 m, centre O, with the rod OP shown at the lowest point, at the point where OP makes an angle of 60 degrees with the downward vertical, and at the highest point; the angle θ from the downward vertical is marked, and the force in the rod is shown acting along OP at the highest point.

(a)Using the work-energy principle, find the speed of P at the point where OP makes an angle of 60 degrees with the downward vertical through O, giving your answer to 3 significant figures.(4)

(b)Find the speed of P at the highest point of the circle, giving your answer to 3 significant figures.(3)

(c)Find the magnitude of the force in the rod when P is at the highest point of the circle, and state whether the rod is in tension or in thrust (compression), ft your answer to (b).(4)

(Total for Question 9 is 11 marks)

A particle P of mass 0.5 kg is attached to one end of a light elastic string of natural length 1.0 m. The other end of the string is attached to a fixed point O. P is held at O and released from rest, falling vertically under gravity. The modulus of elasticity of the string is 20 N. Take g = 9.8 m/s^2.

(a)Find the speed of P at the instant the string first becomes taut.(3)

(b)Given that P first comes to instantaneous rest when the string is stretched by a further extension of e metres beyond its natural length, use energy conservation to show that e satisfies 10*e² - 4.9*e - 4.9 = 0, and hence find e, giving your answer to 3 significant figures.(6)

(c)Hence find the total distance fallen by P when it is first at instantaneous rest.(2)

(Total for Question 10 is 11 marks)

A particle P of mass 0.6 kg is attached to one end of a light elastic string of natural length 0.5 m and modulus of elasticity 36 N. The other end of the string is attached to a fixed point O on a smooth horizontal table. P moves on the table in a horizontal circle of radius 0.65 m, with O at the centre of the circle, and with constant angular speed ω.

(a)Find the extension of the string.(1)

(b)Use Hooke's law to find the tension in the string.(3)

(c)By considering the equation of motion of P towards the centre of the circle, find the value of ω, giving your answer to 3 significant figures.(4)

(d)Find the time taken for P to complete one full revolution, giving your answer to 3 significant figures, ft your answer to (c).(2)

(e)Find the speed of P, giving your answer to 3 significant figures, ft your answer to (c).(2)

(Total for Question 11 is 12 marks)

In a school physics club demonstration in Leeds, run by a technician called Aisha, a particle P of mass 0.2 kg rests in contact with the free end of a light spring of natural length 0.3 m and modulus of elasticity 48 N. The spring is compressed so that its length is 0.15 m, and P is held at rest at the free end of the spring at the bottom of a smooth track. When released, the spring pushes P along a smooth horizontal section of track, after which P moves round the inside of a smooth vertical circular loop of radius 0.35 m. Take g = 9.8 m/s^2.

Figure (to be drawn): Diagram: light spring at the base of the track, then a smooth horizontal section, leading into a vertical circular loop of radius 0.35 m; particle P is shown at the bottom and at the top of the loop, with the normal reaction N marked acting towards the centre at the top.

(a)Find the elastic potential energy (EPE) stored in the spring initially.(3)

(b)Given that all of this elastic potential energy is transferred to kinetic energy of P as it leaves the spring, find the speed of P as it leaves the spring, ft your answer to (a).(2)

(c)By finding the minimum speed at the bottom of the loop needed for P to maintain contact with the track throughout the loop, show that P does complete the loop without losing contact with the track.(4)

(d)Find the speed of P, and the normal reaction of the track on P, at the top of the loop, giving each answer to 3 significant figures.(4)

(Total for Question 12 is 13 marks)

Mark scheme · FP.FM6 Further Mechanics: Work, Energy and Elastic Strings Depth

Question 1

(a) B1 extension = 2.0 - 1.6 = 0.4 m
(a) Answer: 0.4 m
(b) M1 uses equilibrium of P to find the tension: T = mg = 3 x 9.8 = 29.4 (N)
(b) M1 applies Hooke's law T = λ*x/l with their T, x = 0.4 and l = 1.6
(b) A1 λ = 117.6 N, cao
(b) Answer: λ = 117.6 N

Question 2

(a) B1 compression = 0.5 - 0.35 = 0.15 m
(a) Answer: 0.15 m
(b) M1 correct formula EPE = λ*x²/(2*l)
(b) M1 correct substitution: 40*0.15²/(2*0.5)
(b) A1 EPE = 0.9 J, cao
(b) Answer: EPE = 0.9 J
(c) B1 e.g. the spring is light (has negligible mass), or the spring obeys Hooke's law up to this compression, or the spring does not buckle (oe)
(c) Answer: E.g. the spring is light (massless) and obeys Hooke's law up to this compression.

Question 3

(a) M1 finds the normal reaction R = mg = 49 N and the friction force F = μ*R = 0.25*49 = 12.25 N
(a) A1 work done against friction = 12.25*4 = 49 J, cao
(a) Answer: 49 J
(b) M1 finds the work done by the applied force: 30*4 = 120 J
(b) M1 sets up the work-energy equation: (1/2)*m*v² = work by force - work against friction, ft (a)
(b) M1 substitutes to get (1/2)(5)v² = 120 - 49 = 71
(b) A1 v = awrt 5.33 m/s, cao
(b) Answer: v = awrt 5.33 m/s

Question 4

(a) B1 extension = 0.9 - 0.6 = 0.3 m
(a) Answer: 0.3 m
(b) M1 correct formula EPE = λ*x²/(2*l)
(b) M1 correct substitution: 24*0.3²/(2*0.6)
(b) A1 EPE = 1.8 J, cao
(b) Answer: EPE = 1.8 J
(c) M1 uses energy conservation: EPE = KE gained, since the table is smooth and no other horizontal force does work, ft (b)
(c) M1 substitutes: (1/2)(0.4)v² = 1.8
(c) A1 v = 3 m/s, cao
(c) Answer: v = 3 m/s

Question 5

(a) M1 Newton's second law towards the centre (downward) at the top: T + mg = m*v²/r
(a) M1 sets T = 0 for the minimum (critical) case, the string is on the point of becoming slack
(a) A1 v² = gr, cso
(a) A1 v = √9.8*0.75 = awrt 2.71 m/s
(a) Answer: v² = gr; minimum speed = awrt 2.71 m/s
(b) M1 uses energy conservation between top and bottom: (1/2)m*v_bottom² = (1/2)m*v_top² + mg*h
(b) M1 uses the correct height fallen, h = 2r = 1.5 m (the diameter)
(b) M1 substitutes v_top² = gr = 7.35 to get v_bottom² = 7.35 + 2*9.8*1.5 = 36.75, ft (a)
(b) A1 v_bottom = awrt 6.06 m/s, cao
(b) Answer: v_bottom = awrt 6.06 m/s

Question 6

(a) M1 θ = arctan(0.3)
(a) A1 θ = awrt 16.7 degrees
(a) Answer: θ = awrt 16.7 degrees
(b) M1 resolves vertically: N*cos(θ) = mg
(b) M1 resolves horizontally, towards the centre: N*sin(θ) = m*v²/r
(b) A1 divides the two equations to eliminate N and m, giving tan(θ) = v²/(r*g), i.e. v² = r*g*tan(θ), cso
(b) Answer: v² = r*g*tan(θ)
(c) M1 substitutes into v² = r*g*tan(θ): v² = 50*9.8*0.3, ft (b)
(c) A1 v² = 147
(c) A1 v = awrt 12.1 m/s, cao
(c) Answer: v = awrt 12.1 m/s

Question 7

(a) M1 resolves vertically: T*cos(40) = mg
(a) A1 rearranges to T = mg/cos(40), cso
(a) A1 T = awrt 3.84 N
(a) Answer: T = mg/cos(40); T = awrt 3.84 N
(b) M1 uses r = L*sin(40) with L = 0.5 m
(b) A1 r = awrt 0.321 m
(b) Answer: r = awrt 0.321 m
(c) M1 writes the horizontal equation of motion: T*sin(40) = m*ω²*r, ft (a) and (b)
(c) M1 rearranges and substitutes to find ω² = T*sin(40)/(m*r)
(c) A1 ω = awrt 5.06 rad/s
(c) A1 speed v = ω*r = awrt 1.63 m/s, ft their ω and r
(c) Answer: ω = awrt 5.06 rad/s; v = awrt 1.63 m/s

Question 8

(a) M1 correct formula EPE = λ*x²/(2*l) with x = 0.9 - 0.6 = 0.3 m
(a) M1 correct substitution: 90*0.3²/(2*0.9)
(a) A1 EPE = 4.5 J, cao
(a) Answer: EPE = 4.5 J
(b) M1 finds the normal reaction R = mg*cos(30) = awrt 12.7 N
(b) M1 finds the friction force F = μ*R, then work done against friction = F*0.3
(b) A1 work done against friction = awrt 0.764 J
(b) B1 work done against gravity = mg*0.3*sin(30) = 2.205 J
(b) Answer: Work against friction = awrt 0.764 J; work against gravity = 2.205 J
(c) M1 sets up the energy equation: EPE = KE gained + work done against gravity + work done against friction, ft (a) and (b)
(c) M1 substitutes: (1/2)(1.5)v² = 4.5 - 2.205 - 0.764
(c) A1 v = awrt 1.43 m/s, cao
(c) Answer: v = awrt 1.43 m/s

Question 9

(a) M1 finds the height risen above the lowest point: h = r*(1 - cos(60)) = 0.2 m
(a) M1 uses energy conservation: (1/2)*v² = (1/2)*v0² - g*h, i.e. v² = v0² - 2*g*h
(a) M1 substitutes: v² = 16 - 2*9.8*0.2 = 12.08
(a) A1 v = awrt 3.48 m/s, cao
(a) Answer: v = awrt 3.48 m/s
(b) M1 uses the height risen to the top, h = 2r = 0.8 m
(b) M1 substitutes: v_top² = 16 - 2*9.8*0.8 = 0.32
(b) A1 v_top = awrt 0.566 m/s, cao
(b) Answer: v_top = awrt 0.566 m/s
(c) M1 Newton's second law towards the centre (downward) at the top: mg + F = m*v_top²/r, where F is the force in the rod taken as positive towards the centre (tension)
(c) M1 substitutes: F = m*v_top²/r - mg = 0.25*0.32/0.4 - 0.25*9.8, ft (b)
(c) A1 F = -2.25 N, so the magnitude of the force in the rod is 2.25 N
(c) B1 since F is negative (the weight alone exceeds the force needed for circular motion at this speed), the rod pushes P outward, so the rod is in thrust (compression), not tension
(c) Answer: 2.25 N of thrust (the rod is in compression)

Question 10

(a) M1 recognises P falls freely (the string is slack) for the first 1.0 m, and uses v² = u² + 2*g*s (or an energy method) with u = 0, s = 1.0
(a) M1 substitutes: v² = 2*9.8*1.0 = 19.6
(a) A1 v = awrt 4.43 m/s, cao
(a) Answer: v = awrt 4.43 m/s
(b) M1 uses energy conservation between the release point and the lowest point, where both the initial and final kinetic energies are zero: loss in GPE = gain in EPE
(b) M1 writes the total distance fallen as (1.0 + e), giving loss in GPE = mg*(1+e) = 4.9*(1+e)
(b) M1 writes the EPE gained as λ*e²/(2*l) = 20*e²/2 = 10*e²
(b) A1 forms the equation 4.9*(1+e) = 10*e², i.e. 10*e² - 4.9*e - 4.9 = 0, cso
(b) M1 solves the quadratic using the quadratic formula, e = (4.9 ± √4.9² + 4*10*4.9)/20, and rejects the negative root
(b) A1 e = awrt 0.987 m
(b) Answer: 10*e² - 4.9*e - 4.9 = 0; e = awrt 0.987 m
(c) M1 total distance = 1.0 + e, ft (b)
(c) A1 awrt 1.99 m
(c) Answer: awrt 1.99 m

Question 11

(a) B1 extension = 0.65 - 0.5 = 0.15 m
(a) Answer: 0.15 m
(b) M1 correct formula T = λ*x/l
(b) M1 correct substitution: 36*0.15/0.5, ft (a)
(b) A1 T = 10.8 N, cao
(b) Answer: T = 10.8 N
(c) M1 writes the equation of motion towards the centre: T = m*ω²*r, ft (b)
(c) M1 rearranges to ω² = T/(m*r)
(c) M1 substitutes: ω² = 10.8/(0.6*0.65) = 27.7 (awrt)
(c) A1 ω = awrt 5.26 rad/s, cao
(c) Answer: ω = awrt 5.26 rad/s
(d) M1 uses time period = 2*π/ω, ft (c)
(d) A1 awrt 1.19 s
(d) Answer: awrt 1.19 s
(e) M1 uses v = ω*r, ft (c)
(e) A1 awrt 3.42 m/s
(e) Answer: awrt 3.42 m/s

Question 12

(a) M1 correct formula EPE = λ*x²/(2*l) with x = 0.3 - 0.15 = 0.15 m
(a) M1 correct substitution: 48*0.15²/(2*0.3)
(a) A1 EPE = 1.8 J, cao
(a) Answer: EPE = 1.8 J
(b) M1 uses (1/2)*m*v² = EPE, ft (a)
(b) A1 v = √18 = awrt 4.24 m/s, cao
(b) Answer: v = awrt 4.24 m/s
(c) M1 states the critical condition at the top of the loop: the normal reaction N = 0, so v_top² = g*r
(c) M1 uses energy conservation between the bottom and top of the loop (height 2r) to write the corresponding minimum speed at the bottom: v_bottom(min)² = g*r + 2*g*(2r) = 5*g*r
(c) A1 5*g*r = 5*9.8*0.35 = 17.15, so the minimum speed-squared at the bottom is 17.15 (m/s)²
(c) A1 compares with the actual value found in (b), v² = 18 > 17.15, so P has more than enough speed and maintains contact with the track throughout the loop, cso
(c) Answer: Minimum v_bottom² = 17.15; since 18 > 17.15, P maintains contact throughout the loop
(d) M1 uses energy conservation from bottom to top: v_top² = v_bottom² - 2*g*(2r), ft (b)
(d) A1 v_top = awrt 2.07 m/s
(d) M1 Newton's second law towards the centre (downward) at the top: mg + N = m*v_top²/r, ft their v_top
(d) A1 N = awrt 0.486 N, cao
(d) Answer: v_top = awrt 2.07 m/s; N = awrt 0.486 N

Further Mechanics: Work, Energy and Elastic Strings Depth - Worksheets, Questions and Revision