Level 4 — ApplicationWork, Energy & Power

Work, Energy & Power

50 marksprintable — key stays hidden on paper

Level 4: Application (Novel Problems)

Time: 60 minutes Total Marks: 50

Take $g = 9.8\ \text{m/s}^2$ unless otherwise stated. Show all reasoning.

Q1. [10 marks] A block of mass $m = 2.0\ \text{kg}$ is pushed along a horizontal floor by a force whose magnitude varies with position as $F(x) = (12 - 3x)\ \text{N}$ (with $x$ in metres), directed along the motion. The coefficient of kinetic friction between block and floor is $\mu_k = 0.25$ . The block starts from rest at $x = 0$ .

(a) Find the position $x_{\text{stop}}$ where the applied force $F(x)$ first becomes zero. (2) (b) Compute the total work done by the applied force from $x = 0$ to $x = 4.0\ \text{m}$ . (3) (c) Compute the work done by friction over the same interval, and hence the speed of the block at $x = 4.0\ \text{m}$ . (5)

Q2. [12 marks] A ball of mass $m$ is released from rest at the top of a frictionless track that curves down and up. The track's height profile as a function of horizontal position is $h(x) = h_0\,e^{-x/L}$ , so the ball starts at $x=0$ with height $h_0$ . At $x = L$ a rough horizontal patch of length $d$ with kinetic friction coefficient $\mu$ begins.

(a) Using energy conservation, find the speed of the ball as it reaches $x = L$ (where $h = h_0/e$ ). (4) (b) The ball then crosses the rough patch. Derive the condition on $\mu$ , $d$ , $h_0$ , $L$ for the ball to just make it across (arriving with zero speed). (4) (c) With $h_0 = 2.0\ \text{m}$ , $L$ such that $h_0/e = 0.736\ \text{m}$ , $d = 3.0\ \text{m}$ , find the numerical value of $\mu$ for the "just makes it" case. (4)

Q3. [10 marks] A horizontal spring ( $k = 800\ \text{N/m}$ ) is compressed by $0.15\ \text{m}$ and holds a block of mass $0.50\ \text{kg}$ against it. The block is released and slides across a surface with kinetic friction $\mu_k = 0.20$ ; the spring pushes it over its natural-length distance and then the block leaves the spring.

(a) Find the elastic PE stored initially. (2) (b) Find the speed of the block at the instant it leaves the spring (natural length). Assume friction acts throughout the $0.15\ \text{m}$ compression distance. (4) (c) After leaving the spring, how much further does the block slide before stopping? (4)

Q4. [10 marks] A pump raises water from a well $12\ \text{m}$ deep and ejects it at the surface with speed $5.0\ \text{m/s}$ . It delivers $30\ \text{kg}$ of water per second. The pump motor draws electrical power at $6.0\ \text{kW}$ .

(a) Find the mechanical power actually delivered to the water (potential + kinetic energy rate). (5) (b) Compute the efficiency of the pump. (3) (c) State two physical reasons why real efficiency is below 100%. (2)

Q5. [8 marks] A force acting in the $xy$ -plane is $\vec{F} = (2y)\,\hat{i} + (x)\,\hat{j}$ (SI units).

(a) Compute the work done by $\vec{F}$ moving a particle from $(0,0)$ to $(1,1)$ along path A: first along the $x$ -axis to $(1,0)$ , then vertically to $(1,1)$ . (3) (b) Compute the work along path B: the straight line $y = x$ from $(0,0)$ to $(1,1)$ . (3) (c) State whether $\vec{F}$ is conservative and justify using your results. (2)

Answer keyMark scheme & solutions

Q1

(a) $F(x)=12-3x=0 \Rightarrow x_{\text{stop}} = 4.0\ \text{m}$ . (2)

(b) $W_F=\int_0^4 (12-3x)\,dx = [12x - 1.5x^2]_0^4 = 48 - 24 = 24\ \text{J}$ . (3)

(c) Friction force $f=\mu_k m g = 0.25\times 2.0\times 9.8 = 4.9\ \text{N}$ . $W_{fric} = -f\cdot d = -4.9\times 4.0 = -19.6\ \text{J}$ . (2) Work-energy theorem: $\tfrac12 m v^2 = W_F + W_{fric} = 24 - 19.6 = 4.4\ \text{J}$ . (2) $v = \sqrt{2\times 4.4/2.0} = \sqrt{4.4} = 2.10\ \text{m/s}$ . (1)

Q2

(a) Height drop from $h_0$ to $h_0/e$ : $\Delta h = h_0(1-1/e)$ . $\tfrac12 m v^2 = mg\,h_0(1-1/e) \Rightarrow v = \sqrt{2 g h_0 (1-1/e)}$ . (4)

(b) Kinetic energy at $x=L$ must be dissipated by friction over $d$ : $\tfrac12 m v^2 = \mu m g d \Rightarrow g h_0(1-1/e) = \mu g d$ $\boxed{\mu = \dfrac{h_0(1-1/e)}{d}}$ . (4)

(c) $h_0(1-1/e) = 2.0 - 0.736 = 1.264\ \text{m}$ . $\mu = 1.264/3.0 = 0.421$ . (4)

Q3

(a) $U = \tfrac12 k x^2 = \tfrac12 (800)(0.15)^2 = 400\times 0.0225 = 9.0\ \text{J}$ . (2)

(b) Friction force $f=\mu_k m g = 0.20\times 0.50\times 9.8 = 0.98\ \text{N}$ . Work by friction over $0.15\ \text{m}$ : $W_f = -0.98\times 0.15 = -0.147\ \text{J}$ . (2) Energy: $\tfrac12 m v^2 = U - |W_f| = 9.0 - 0.147 = 8.853\ \text{J}$ . $v = \sqrt{2\times 8.853/0.50} = \sqrt{35.41} = 5.95\ \text{m/s}$ . (2)

(c) After leaving spring, KE $=8.853\ \text{J}$ dissipated by friction: $8.853 = f\cdot s = 0.98\, s \Rightarrow s = 9.03\ \text{m}$ . (4)

Q4

(a) Mass rate $\dot m = 30\ \text{kg/s}$ . PE rate $= \dot m g h = 30\times 9.8\times 12 = 3528\ \text{W}$ . KE rate $= \tfrac12 \dot m v^2 = \tfrac12\times 30\times 25 = 375\ \text{W}$ . Total useful power $= 3528 + 375 = 3903\ \text{W} \approx 3.90\ \text{kW}$ . (5)

(b) $\eta = 3903/6000 = 0.6505 \approx 65\%$ . (3)

(c) Any two: heat from friction in bearings/pipes; viscous/turbulent losses in water flow; electrical resistive (I²R) heating in motor windings; sound/vibration losses. (2)

Q5

(a) Path A. Segment 1 ( $y=0$ , $dy=0$ ): $W_1=\int_0^1 2y\,dx = 0$ . Segment 2 ( $x=1$ , $dx=0$ ): $W_2=\int_0^1 x\,dy = \int_0^1 1\,dy = 1$ . $W_A = 0 + 1 = 1\ \text{J}$ . (3)

(b) Path B: $y=x$ , $dy=dx$ . $W_B = \int_0^1 (2y\,dx + x\,dy) = \int_0^1 (2x + x)\,dx = \int_0^1 3x\,dx = 1.5\ \text{J}$ . (3)

(c) $W_A \neq W_B$ (path-dependent) $\Rightarrow \vec F$ is non-conservative. (Check: $\partial F_x/\partial y = 2 \neq \partial F_y/\partial x = 1$ .) (2)

[
{"claim":"Q1c: v at x=4 is sqrt(4.4)=2.098 m/s","code":"WF=integrate(12-3*x,(x,0,4)); f=0.25*2.0*9.8; KE=WF-f*4.0; v=sqrt(2*KE/2.0); result=abs(float(v)-2.0976)<0.01"},
{"claim":"Q2c: mu = 0.421","code":"h0=Rational(2); term=h0*(1-1/E); mu=term/3; result=abs(float(mu)-0.4213)<0.005"},
{"claim":"Q3: U=9.0J, v=5.95 m/s, s=9.03 m","code":"U=Rational(1,2)*800*Rational(15,100)**2; f=0.20*0.50*9.8; KE=float(U)-f*0.15; v=sqrt(2*KE/0.50); s=KE/f; result=(abs(float(U)-9.0)<0.01) and (abs(float(v)-5.951)<0.02) and (abs(s-9.033)<0.05)"},
{"claim":"Q4: useful power 3903 W, eff 0.65","code":"P=30*9.8*12+0.5*30*25; eta=P/6000; result=(abs(P-3903)<1) and (abs(eta-0.6505)<0.005)"},
{"claim":"Q5: WA=1, WB=1.5, non-conservative","code":"WA=0+integrate(1,(y,0,1)); WB=integrate(3*x,(x,0,1)); result=(WA==1) and (WB==Rational(3,2)) and (WA!=WB)"}
]