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Work and Power

Calculating work using line integrals and instantaneous power

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Work and Power

Work as a Line Integral

For a variable force F⃗\vec{F} along a path from point A to B:

W=∫ABF⃗⋅dr⃗W = \int_A^B \vec{F} \cdot d\vec{r}

In one dimension: W=∫x1x2Fx dxW = \int_{x_1}^{x_2} F_x \, dx

In three dimensions: W=∫AB(Fx dx+Fy dy+Fz dz)W = \int_A^B (F_x \, dx + F_y \, dy + F_z \, dz)

Work-Energy Theorem

From Newton's second law: F=mdvdt=mdvdxdxdt=mvdvdxF = m\frac{dv}{dt} = m\frac{dv}{dx}\frac{dx}{dt} = mv\frac{dv}{dx}

F dx=mv dvF \, dx = mv \, dv

Integrating: ∫x1x2F dx=∫v1v2mv dv\int_{x_1}^{x_2} F \, dx = \int_{v_1}^{v_2} mv \, dv

W=12mv22−12mv12=ΔKEW = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2 = \Delta KE

Work-energy theorem: Wnet=ΔKEW_{net} = \Delta KE

Common Force Examples

Constant Force

W=F⋅d=Fdcos⁡θW = F \cdot d = Fd\cos\theta

where θ\theta is angle between force and displacement.

Spring Force

F=−kxF = -kx

W=∫0x(−kx′) dx′=−12kx2W = \int_0^x (-kx') \, dx' = -\frac{1}{2}kx^2

(Work done by spring; work done on spring is positive)

Gravity Near Earth's Surface

W=∫y1y2(−mg) dy=−mg(y2−y1)=−mgΔhW = \int_{y_1}^{y_2} (-mg) \, dy = -mg(y_2 - y_1) = -mg\Delta h

Inverse Square Force (Gravity/Electrostatic)

F=−kr2F = -\frac{k}{r^2}

W=∫r1r2(−kr2)dr=k(1r2−1r1)W = \int_{r_1}^{r_2} \left(-\frac{k}{r^2}\right) dr = k\left(\frac{1}{r_2} - \frac{1}{r_1}\right)

Power

Instantaneous power: P=dWdtP = \frac{dW}{dt}

Since dW=F⃗⋅dr⃗dW = \vec{F} \cdot d\vec{r}:

P=dWdt=F⃗⋅dr⃗dt=F⃗⋅v⃗P = \frac{dW}{dt} = \vec{F} \cdot \frac{d\vec{r}}{dt} = \vec{F} \cdot \vec{v}

In one dimension: P=FvP = Fv

Average Power

Pavg=WΔtP_{avg} = \frac{W}{\Delta t}

Power and Kinetic Energy

P=dKEdt=ddt(12mv2)=mvdvdt=mva=FvP = \frac{dKE}{dt} = \frac{d}{dt}\left(\frac{1}{2}mv^2\right) = mv\frac{dv}{dt} = mva = Fv

Work on Variable Mass Systems

For rocket with exhaust velocity vev_e relative to rocket:

Power delivered by engine: P=Fthrust⋅v=vrel∣dmdt∣⋅vP = F_{thrust} \cdot v = v_{rel}\left|\frac{dm}{dt}\right| \cdot v

Example: Work Against Drag

Object moving through fluid with drag Fd=−bv2F_d = -bv^2:

W=∫0dFd dx=−b∫0dv2 dxW = \int_0^d F_d \, dx = -b\int_0^d v^2 \, dx

If v(t)v(t) is known, use dx=v dtdx = v \, dt:

W=−b∫0Tv3 dtW = -b\int_0^T v^3 \, dt

This work is dissipated as heat.

📚 Practice Problems

1Problem 1medium

❓ Question:

A variable force F(x) = (10 - 2x) N acts on a 3.0 kg object, where x is in meters. The object starts from rest at x = 0. Find: (a) the work done from x = 0 to x = 4 m, (b) the final speed at x = 4 m, and (c) the power delivered at x = 2 m.

💡 Show Solution

Given:

  • F(x) = 10 - 2x N
  • m = 3.0 kg
  • x₀ = 0, v₀ = 0

(a) Work done from x = 0 to x = 4 m:

W=∫04F(x) dx=∫04(10−2x) dxW = \int_0^4 F(x) \, dx = \int_0^4 (10 - 2x) \, dx

W=[10x−x2]04=40−16W = \left[10x - x^2\right]_0^4 = 40 - 16

W=24 J\boxed{W = 24 \text{ J}}

(b) Final speed at x = 4 m:

Using work-energy theorem: W=ΔKE=12mv2−0W = \Delta KE = \frac{1}{2}mv^2 - 0

24=12(3.0)v224 = \frac{1}{2}(3.0)v^2

v2=16v^2 = 16

v=4.0 m/s\boxed{v = 4.0 \text{ m/s}}

(c) Power at x = 2 m:

First find velocity at x = 2: W0→2=∫02(10−2x) dx=20−4=16 JW_{0→2} = \int_0^2 (10 - 2x) \, dx = 20 - 4 = 16 \text{ J}

12(3.0)v22=16  ⟹  v2=3.27 m/s\frac{1}{2}(3.0)v_2^2 = 16 \implies v_2 = 3.27 \text{ m/s}

Force at x = 2: F(2)=10−2(2)=6 NF(2) = 10 - 2(2) = 6 \text{ N}

Power: P=F⋅v=(6)(3.27)P = F \cdot v = (6)(3.27)

P=19.6 W\boxed{P = 19.6 \text{ W}}

2Problem 2hard

❓ Question:

A car of mass m = 1200 kg accelerates from rest. Its engine provides constant power P = 60 kW. Neglecting friction, find: (a) the velocity as a function of time, (b) the time to reach v = 25 m/s, and (c) the distance traveled in this time.

💡 Show Solution

Given:

  • m = 1200 kg
  • P = 60 kW = 60,000 W
  • v₀ = 0

(a) Velocity as function of time:

P=Fv=mav=mvdvdtP = Fv = mav = mv\frac{dv}{dt}

dvdt=Pmv\frac{dv}{dt} = \frac{P}{mv}

Separating variables: v dv=Pmdtv \, dv = \frac{P}{m}dt

∫0vv′ dv′=∫0tPmdt′\int_0^v v' \, dv' = \int_0^t \frac{P}{m} dt'

v22=Pmt\frac{v^2}{2} = \frac{P}{m}t

v2=2Ptmv^2 = \frac{2Pt}{m}

v(t)=2Ptm=2(60000)t1200=100t=10t m/s\boxed{v(t) = \sqrt{\frac{2Pt}{m}} = \sqrt{\frac{2(60000)t}{1200}} = \sqrt{100t} = 10\sqrt{t} \text{ m/s}}

(b) Time to reach 25 m/s:

25=10t25 = 10\sqrt{t}

t=2.5\sqrt{t} = 2.5

t=6.25 s\boxed{t = 6.25 \text{ s}}

(c) Distance traveled:

v=dxdt=10tv = \frac{dx}{dt} = 10\sqrt{t}

dx=10t dtdx = 10\sqrt{t} \, dt

x=∫06.2510t dt=10⋅2t3/23∣06.25x = \int_0^{6.25} 10\sqrt{t} \, dt = 10 \cdot \frac{2t^{3/2}}{3}\Big|_0^{6.25}

x=203(6.25)3/2=203(15.625)x = \frac{20}{3}(6.25)^{3/2} = \frac{20}{3}(15.625)

x=104 m\boxed{x = 104 \text{ m}}

3Problem 3easy

❓ Question:

A pump lifts water from a well of depth h = 20 m at a rate of dm/dt = 15 kg/s. Find: (a) the power required (ignoring kinetic energy of water), (b) if the pump is 75% efficient, what is the actual power input needed? (c) How much water is lifted in 5 minutes?

💡 Show Solution

Given:

  • h = 20 m
  • dm/dt = 15 kg/s
  • g = 9.8 m/s²
  • Efficiency = 75%

(a) Power required:

Rate of change of potential energy: P=dUdt=d(mgh)dt=ghdmdtP = \frac{dU}{dt} = \frac{d(mgh)}{dt} = gh\frac{dm}{dt}

P=(9.8)(20)(15)P = (9.8)(20)(15)

P=2940 W=2.94 kW\boxed{P = 2940 \text{ W} = 2.94 \text{ kW}}

(b) Actual power input:

Pinput=Poutputη=29400.75P_{input} = \frac{P_{output}}{\eta} = \frac{2940}{0.75}

Pinput=3920 W=3.92 kW\boxed{P_{input} = 3920 \text{ W} = 3.92 \text{ kW}}

(c) Water lifted in 5 minutes:

Δm=dmdt⋅Δt=(15)(5×60)\Delta m = \frac{dm}{dt} \cdot \Delta t = (15)(5 \times 60)

Δm=4500 kg\boxed{\Delta m = 4500 \text{ kg}}

Volume (ρ = 1000 kg/m³): V=mρ=45001000=4.5 m3=4500 LV = \frac{m}{\rho} = \frac{4500}{1000} = 4.5 \text{ m}^3 = 4500 \text{ L}

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Calculating work using line integrals and instantaneous power
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