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PhyMedAdv 2019 • Paper 1

Work Done by Non Conservative Force along Closed Trajectory

A particle is moved along the closed path AB-BC-CD-DE-EF-FA, as depicted in the figure, in the presence of a force F⃗=(αyi^+2αxj^) N\vec{F} = (\alpha y\hat{i} + 2\alpha x\hat{j})\text{ N}, where xx and yy are in meters and α=−1 Nm−1\alpha = -1\text{ Nm}^{-1}. The total work performed on the particle by this force F⃗\vec{F} is _____ Joule.

Question Diagram 1
Official Numerical Answer0.75

Topics & Concepts

Work, Energy and PowerWork-Energy Theorem

Step-by-Step Solution

To find the total work done on the particle by the force F⃗=(αyi^+2αxj^) N\vec{F} = (\alpha y\hat{i} + 2\alpha x\hat{j})\text{ N} along the closed path AB−BC−CD−DE−EF−FAAB-BC-CD-DE-EF-FA, we evaluate the line integral:

W=∮F⃗⋅dr⃗=∮(Fxdx+Fydy)=∮(αydx+2αxdy)W = \oint \vec{F} \cdot d\vec{r} = \oint (F_x dx + F_y dy) = \oint (\alpha y dx + 2\alpha x dy)

Given parameters:

  • α=−1 N m−1\alpha = -1\text{ N m}^{-1}
  • Path coordinates:
    • A(0,1.0)A(0, 1.0)
    • B(1.0,1.0)B(1.0, 1.0)
    • C(1.0,0.5)C(1.0, 0.5)
    • D(0.5,0.5)D(0.5, 0.5)
    • E(0.5,0)E(0.5, 0)
    • F(0,0)F(0, 0)

We can calculate the work done segment by segment along the path:

  1. Segment AB: From (0,1.0)(0, 1.0) to (1.0,1.0)(1.0, 1.0)

    • y=1.0  ⟹  dy=0y = 1.0 \implies dy = 0, xx goes from 00 to 1.01.0.
    • WAB=∫01.0α(1.0)dx=α[x]01.0=α=−1 JW_{AB} = \int_{0}^{1.0} \alpha (1.0) dx = \alpha [x]_0^{1.0} = \alpha = -1\text{ J}

  2. Segment BC: From (1.0,1.0)(1.0, 1.0) to (1.0,0.5)(1.0, 0.5)

    • x=1.0  ⟹  dx=0x = 1.0 \implies dx = 0, yy goes from 1.01.0 to 0.50.5.
    • WBC=∫1.00.52α(1.0)dy=2α[y]1.00.5=2α(0.5−1.0)=−α=1 JW_{BC} = \int_{1.0}^{0.5} 2\alpha (1.0) dy = 2\alpha [y]_{1.0}^{0.5} = 2\alpha (0.5 - 1.0) = -\alpha = 1\text{ J}

  3. Segment CD: From (1.0,0.5)(1.0, 0.5) to (0.5,0.5)(0.5, 0.5)

    • y=0.5  ⟹  dy=0y = 0.5 \implies dy = 0, xx goes from 1.01.0 to 0.50.5.
    • WCD=∫1.00.5α(0.5)dx=0.5α[x]1.00.5=0.5α(0.5−1.0)=−0.25α=0.25 JW_{CD} = \int_{1.0}^{0.5} \alpha (0.5) dx = 0.5\alpha [x]_{1.0}^{0.5} = 0.5\alpha (0.5 - 1.0) = -0.25\alpha = 0.25\text{ J}

  4. Segment DE: From (0.5,0.5)(0.5, 0.5) to (0.5,0)(0.5, 0)

    • x=0.5  ⟹  dx=0x = 0.5 \implies dx = 0, yy goes from 0.50.5 to 00.
    • WDE=∫0.502α(0.5)dy=α[y]0.50=α(0−0.5)=−0.5α=0.5 JW_{DE} = \int_{0.5}^{0} 2\alpha (0.5) dy = \alpha [y]_{0.5}^{0} = \alpha (0 - 0.5) = -0.5\alpha = 0.5\text{ J}

  5. Segment EF: From (0.5,0)(0.5, 0) to (0,0)(0, 0)

    • y=0  ⟹  dy=0y = 0 \implies dy = 0
    • WEF=∫0.50α(0)dx=0 JW_{EF} = \int_{0.5}^{0} \alpha (0) dx = 0\text{ J}

  6. Segment FA: From (0,0)(0, 0) to (0,1.0)(0, 1.0)

    • x=0  ⟹  dx=0x = 0 \implies dx = 0
    • WFA=∫01.02α(0)dy=0 JW_{FA} = \int_{0}^{1.0} 2\alpha (0) dy = 0\text{ J}

Adding the work done along all individual segments gives the total work done WW:

W=WAB+WBC+WCD+WDE+WEF+WFAW = W_{AB} + W_{BC} + W_{CD} + W_{DE} + W_{EF} + W_{FA} W=−1+1+0.25+0.5+0+0=0.75 JW = -1 + 1 + 0.25 + 0.5 + 0 + 0 = 0.75\text{ J}

(Alternatively, using Green's Theorem for a clockwise oriented closed curve CC enclosing region RR: W=−∬R(∂Fy∂x−∂Fx∂y)dA=−∬R(2α−α)dA=−αArea(R)W = -\iint_R \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right) dA = -\iint_R (2\alpha - \alpha) dA = -\alpha \text{Area}(R) Area(R)=(0.5×1.0)+(0.5×0.5)=0.75 m2\text{Area}(R) = (0.5 \times 1.0) + (0.5 \times 0.5) = 0.75\text{ m}^2 W=−(−1)×0.75=0.75 J)W = -(-1) \times 0.75 = 0.75\text{ J}\text{)}

Final Answer: The total work performed on the particle by the force F⃗\vec{F} is 0.75 Joule.

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