Noesis
Level 2 — RecallWork, Energy & Power

Work, Energy & Power

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Level 2 (Recall & Standard Problems)

Time: 30 minutes | Total Marks: 40

Take g=10 m s−2g = 10\ \text{m s}^{-2} unless stated otherwise.

Q1. State the definition of work done by a constant force. Write it as a dot product and state when work is (a) positive, (b) negative, (c) zero. (3 marks)

Q2. A force F⃗=(3i^+4j^) N\vec{F} = (3\hat{i} + 4\hat{j})\ \text{N} moves a particle through displacement d⃗=(5i^−2j^) m\vec{d} = (5\hat{i} - 2\hat{j})\ \text{m}. Calculate the work done. (3 marks)

Q3. A variable force F(x)=6x2 NF(x) = 6x^2\ \text{N} acts on a body along the xx-axis. Find the work done in moving it from x=1 mx = 1\ \text{m} to x=3 mx = 3\ \text{m}. (4 marks)

Q4. State the work–energy theorem and derive it starting from Newton's second law for a constant force acting over displacement ss. (5 marks)

Q5. A 2 kg2\ \text{kg} ball is dropped from rest at a height of 5 m5\ \text{m}. Using conservation of mechanical energy, find its speed just before hitting the ground. (4 marks)

Q6. Derive the expression for the elastic potential energy stored in a spring of stiffness kk stretched by a displacement xx, starting from Hooke's law. (5 marks)

Q7. A block of mass 4 kg4\ \text{kg} slides 6 m6\ \text{m} across a rough horizontal surface with coefficient of kinetic friction μ=0.25\mu = 0.25. Find the work done by friction and state its sign. (4 marks)

Q8. A pump lifts 600 kg600\ \text{kg} of water through a height of 20 m20\ \text{m} in 40 s40\ \text{s}. Calculate the useful power output. If the input electrical power is 4 kW4\ \text{kW}, find the efficiency. (5 marks)

Q9. A car of mass 1000 kg1000\ \text{kg} moving at 20 m s−120\ \text{m s}^{-1} compresses a spring bumper (k=5×105 N m−1k = 5 \times 10^{5}\ \text{N m}^{-1}) on impact. Assuming all kinetic energy converts to spring PE, find the maximum compression. (4 marks)

Q10. Distinguish between conservative and non-conservative forces, giving one example of each. (3 marks)

End of paper

Answer keyMark scheme & solutions

Q1. (3 marks) Work =F⃗⋅d⃗=Fdcos⁡θ= \vec{F}\cdot\vec{d} = Fd\cos\theta, where θ\theta is angle between force and displacement. (1)

  • Positive when 0°≤θ<90°0° \le \theta < 90° (force has component along motion). (½)
  • Negative when 90°<θ≤180°90° < \theta \le 180°. (½)
  • Zero when θ=90°\theta = 90° (force perpendicular to displacement) or d=0d=0. (1)

Q2. (3 marks) W=F⃗⋅d⃗=(3)(5)+(4)(−2)W = \vec{F}\cdot\vec{d} = (3)(5) + (4)(-2) (1) =15−8= 15 - 8 (1) =7 J= 7\ \text{J} (1)

Q3. (4 marks) W=∫13F dx=∫136x2 dxW = \int_{1}^{3} F\,dx = \int_{1}^{3} 6x^2\,dx (1) =[2x3]13= \left[2x^3\right]_{1}^{3} (1) =2(27)−2(1)=54−2= 2(27) - 2(1) = 54 - 2 (1) =52 J= 52\ \text{J} (1)

Q4. (5 marks) Statement: Net work done on a body equals the change in its kinetic energy, Wnet=ΔKEW_{net} = \Delta KE. (1) Derivation: From F=maF = ma, and v2=u2+2asv^2 = u^2 + 2as (1) ⇒as=v2−u22\Rightarrow as = \dfrac{v^2 - u^2}{2}. Multiply Newton's law by ss: (1) Fs=mas=m(v2−u22)Fs = mas = m\left(\dfrac{v^2-u^2}{2}\right) (1) ⇒W=12mv2−12mu2=ΔKE\Rightarrow W = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mu^2 = \Delta KE. (1)

Q5. (4 marks) Conservation: mgh=12mv2mgh = \tfrac{1}{2}mv^2 (1) v=2ghv = \sqrt{2gh} (1) =2×10×5=100= \sqrt{2 \times 10 \times 5} = \sqrt{100} (1) =10 m s−1= 10\ \text{m s}^{-1} (1)

Q6. (5 marks) Hooke's law: applied force to stretch =kx= kx (restoring force F=−kxF=-kx). (1) Work done stretching from 00 to xx against spring: (1) U=∫0xkx′ dx′U = \int_{0}^{x} kx'\,dx' (1) =k[x′22]0x= k\left[\dfrac{x'^2}{2}\right]_{0}^{x} (1) =12kx2= \tfrac{1}{2}kx^2. This equals stored elastic PE. (1)

Q7. (4 marks) Friction force f=μmg=0.25×4×10=10 Nf = \mu mg = 0.25 \times 4 \times 10 = 10\ \text{N} (2) Wf=−f⋅s=−10×6=−60 JW_f = -f\cdot s = -10 \times 6 = -60\ \text{J} (1) Negative because friction opposes motion. (1)

Q8. (5 marks) Useful work =mgh=600×10×20=120000 J= mgh = 600 \times 10 \times 20 = 120000\ \text{J} (1) Useful power P=Wt=12000040=3000 W=3 kWP = \dfrac{W}{t} = \dfrac{120000}{40} = 3000\ \text{W} = 3\ \text{kW} (2) Efficiency =PoutPin=30004000= \dfrac{P_{out}}{P_{in}} = \dfrac{3000}{4000} (1) =0.75=75%= 0.75 = 75\% (1)

Q9. (4 marks) Energy: 12mv2=12kx2\tfrac{1}{2}mv^2 = \tfrac{1}{2}kx^2 (1) x=vmk=2010005×105x = v\sqrt{\dfrac{m}{k}} = 20\sqrt{\dfrac{1000}{5\times10^5}} (1) =202×10−3=20×0.04472= 20\sqrt{2\times10^{-3}} = 20 \times 0.04472 (1) ≈0.894 m\approx 0.894\ \text{m} (1)

Q10. (3 marks)

  • Conservative force: work done is path-independent; depends only on endpoints; a PE can be defined; e.g. gravity, spring force. (1½)
  • Non-conservative force: work depends on path; no PE definable; mechanical energy not conserved; e.g. friction, air drag. (1½)
[
  {"claim":"Q2 work equals 7 J","code":"F=Matrix([3,4]); d=Matrix([5,-2]); result=(F.dot(d)==7)"},
  {"claim":"Q3 integral of 6x^2 from 1 to 3 is 52","code":"x=symbols('x'); result=(integrate(6*x**2,(x,1,3))==52)"},
  {"claim":"Q5 speed sqrt(2gh)=10","code":"result=(sqrt(2*10*5)==10)"},
  {"claim":"Q8 efficiency is 0.75","code":"P_out=600*10*20/40; result=(Rational(P_out,4000)==Rational(3,4))"},
  {"claim":"Q9 max compression approx 0.894 m","code":"x=20*sqrt(Rational(1000,500000)); result=(abs(float(x)-0.894427)<1e-3)"}
]