Work, Energy and Power: Edge Cases and Subtle Traps (9)

hard 2 min read

Question

A particle moves along the xx-axis under the variable force F(x)=(3x22x) NF(x) = (3x^2 - 2x)\text{ N}, where xx is in metres. Find the work done by this force as the particle moves from x=1 mx = 1\text{ m} to x=3 mx = 3\text{ m}. Then, if the particle has mass m=2 kgm = 2\text{ kg} and starts from rest at x=1x = 1, find its speed at x=3x = 3.

Solution — Step by Step

Force is variable — it depends on xx. So W=FsW = F \cdot s does not apply. We have to use the integral definition: W=FdxW = \int F\,dx. This is the trap of the “work” chapter — students see “force × distance” and forget the formula breaks for non-constant FF.

W=13(3x22x)dx=[x3x2]13W = \int_{1}^{3} (3x^2 - 2x)\,dx = \left[x^3 - x^2\right]_{1}^{3} W=(279)(11)=18 JW = (27 - 9) - (1 - 1) = 18\text{ J}

Wnet=ΔKE=12mv20W_{\text{net}} = \Delta KE = \tfrac{1}{2}mv^2 - 0. So:

18=12(2)v2    v2=18    v=18=32 m/s18 = \tfrac{1}{2}(2)v^2 \implies v^2 = 18 \implies v = \sqrt{18} = 3\sqrt{2}\text{ m/s}

W=18 JW = 18\text{ J}, v=324.24 m/sv = 3\sqrt{2} \approx 4.24\text{ m/s}.

Why This Works

The work-energy theorem holds for any force — constant, variable, or even non-conservative. As long as we compute net work correctly (here, only one force acts, so net work = work by FF), the change in KE follows directly.

Integration replaces multiplication when the integrand varies. We are summing FdxF\,dx over infinitesimal slices of the path.

Alternative Method

If FF were a conservative force, we could find a potential energy U(x)U(x) such that F=dU/dxF = -dU/dx, and compute W=ΔUW = -\Delta U. Here U(x)=x3+x2U(x) = -x^3 + x^2 (up to a constant), giving W=[U(3)U(1)]=[(27+9)(1+1)]=18 JW = -[U(3) - U(1)] = -[(-27+9) - (-1+1)] = 18\text{ J}. Same answer — useful when you have to track energy across multiple stages.

Common Mistake

Two trap variants. (1) Plugging the average force Fˉ=(F(1)+F(3))/2\bar F = (F(1) + F(3))/2 and multiplying by Δx\Delta x — this works only for linear F(x)F(x), not quadratic. (2) Forgetting that the particle starts from rest; the theorem says ΔKE\Delta KE, not KEfinalKE_{\text{final}}. If v00v_0 \ne 0, you must add the initial KE.

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