Work, Energy & Power

Q1 (10)

(a) [6] Newton's second law: $F = ma = m\dfrac{dv}{dt}$. [1]

Multiply by $dx$ and integrate over the path: $W_{net}=\int_{x_i}^{x_f} F\,dx = \int m\frac{dv}{dt}\,dx$ [1]

Change variable using $\dfrac{dv}{dt} = \dfrac{dv}{dx}\dfrac{dx}{dt} = v\dfrac{dv}{dx}$ [2] (chain rule / $dx = v\,dt$)

$W_{net} = \int_{v_i}^{v_f} m v\,dv = \tfrac12 m v_f^2 - \tfrac12 m v_i^2$ [2]

(b) [4]

• The $\tfrac12 m v^2$ form arises because the integrand became $mv\,dv$ after the change of variable. [2]
• Integrating $v\,dv$ gives $\tfrac12 v^2$ — the power and coefficient are fixed by the mathematics of the integral, not chosen. The chain-rule substitution is the essential step that produces exactly this function. [2]

Q2 (9)

(a) [3] $F = -kx$. [1] The minus sign means the spring force is a restoring force [1], always directed opposite to the displacement from equilibrium (toward equilibrium). [1]

(b) [4] $U(x) = -\int_0^x F(x')dx' = -\int_0^x(-kx')dx'$ [2] $= k\int_0^x x'\,dx' = k\cdot\frac{x^2}{2} = \tfrac12 kx^2$ [2]

(c) [2] $W = U(0.20)-U(0.10) = \tfrac12(200)(0.20^2 - 0.10^2)$ $= 100(0.04-0.01) = 100(0.03) = 3.0\ \text{J}$ [2]

Q3 (10)

(a) [2] A force is conservative if the work it does on a particle moving between two points is independent of the path taken (equivalently, work around any closed loop is zero). [2]

(b) [5] Work–energy theorem: $W_{net} = \Delta K$. [1] With only conservative forces, $W_{net} = W_{cons} = -\Delta U$. [2] So $\Delta K = -\Delta U \Rightarrow \Delta K + \Delta U = 0$. [1] Hence $\Delta(K+U)=0 \Rightarrow E = K+U = \text{const}$. [1]

(c) [3] $mgh = \tfrac12 m v^2 \Rightarrow v = \sqrt{2gh}$ [1] $v = \sqrt{2(9.8)(5.0)} = \sqrt{98} = 9.90\ \text{m/s}$ [2] (mass cancels)

Q4 (9)

(a) [4] Work–energy theorem including all forces: $W_{engine} - W_{diss} = \Delta K$ [1] $\Delta K = \tfrac12(1500)(20^2) = 3.0\times10^5\ \text{J}$ [1] $W_{engine} = \Delta K + W_{diss} = 3.0\times10^5 + 1.2\times10^5 = 4.2\times10^5\ \text{J}$ [2]

(b) [3] Useful power = $\Delta K/t = 3.0\times10^5/8.0 = 3.75\times10^4\ \text{W}$ [1.5] Total engine power = $W_{engine}/t = 4.2\times10^5/8.0 = 5.25\times10^4\ \text{W}$ [1.5]

(c) [2] $\eta = \frac{\text{useful KE gain}}{\text{total engine work}} = \frac{3.0\times10^5}{4.2\times10^5} = 0.714 = 71.4\%$ [2]

Q5 (8)

(a) [4] At max compression the block is momentarily at rest; all KE stored as spring PE: $\tfrac12 m v^2 = \tfrac12 k x_{max}^2$ [2] $x_{max} = v\sqrt{\frac{m}{k}} = 3.0\sqrt{\frac{0.50}{800}}$ [1] $= 3.0\sqrt{6.25\times10^{-4}} = 3.0(0.025) = 0.075\ \text{m}$ [1]

(b) [4]

• The block's speed is maximum at the instant it first contacts the spring (equilibrium/natural length), i.e. before any compression. [2]
• Because the spring force opposes motion throughout compression, it decelerates the block from the moment of contact; speed only decreases during compression. Maximum compression is where speed = 0 (all KE converted). The two points are distinct: max speed = zero net decelerating force (start of contact), max compression = zero speed. [2]

Q6 (4)

$W = \int_1^3 (6x^2 - 4x)\,dx = \left[2x^3 - 2x^2\right]_1^3$ [2] $= (2\cdot27 - 2\cdot9) - (2 - 2) = (54-18) - 0 = 36\ \text{J}$ [2]

[
  {"claim":"Q2c work to stretch spring = 3.0 J","code":"k=200; result = (0.5*k*(0.20**2-0.10**2))==3.0"},
  {"claim":"Q3c speed at bottom = sqrt(98)","code":"result = simplify((2*Rational(98,10)*5) - 98)==0 and abs(float(sqrt(98))-9.8995)<1e-3"},
  {"claim":"Q4 engine work 4.2e5 J and efficiency 5/7","code":"dK=Rational(1,2)*1500*400; Weng=dK+120000; result = Weng==420000 and simplify(dK/Weng-Rational(5,7))==0"},
  {"claim":"Q5a max compression = 0.075 m","code":"result = simplify(3.0*sqrt(Rational(1,2)/800) - 0.075)==0"},
  {"claim":"Q6 work integral of 6x^2-4x from 1 to 3 = 36 J","code":"x=symbols('x'); result = integrate(6*x**2-4*x,(x,1,3))==36"}
]