Question

A force acting on a particle moving in the x-y plane is given by F=2yi+x^2j N, where x and y are in meters. The particle moves from the origin to a final position having the coordinates x=5 m and y=5 m and shown in the figure above. Calculate the work done by F along (a) OAC, (b) OBC, and (c) OC. (d) Is F a conservative or non-conservative force? Explain?

Answer 1

Given that the force \(\mathrm{F}=2 \mathrm{y} \mathrm{i}+\left(x^{2}\right)\)

Work done along path OAis

$$ \begin{aligned} \mathrm{W}_{0 \mathrm{~A}} &=\int F d x+\int F \cdot d y+\int F \cdot d z \\ &=0+0+0=0 \end{aligned} $$

Work done along path \(A C\) is

$$ \begin{aligned} \mathrm{W}_{A C} &=\int_{0}^{0} F \cdot d x+\int_{0}^{5} F \cdot d y+\int_{0}^{0} F \cdot d z \\ &=0+\int_{0}^{5}\left(x^{2}\right) \cdot d y+0 \\ &=\int_{0}^{5}(25) \cdot d y \\ &=25 \times 5=125 \mathrm{~J} \end{aligned} $$

The work done along \(\mathrm{OAC}\) is \(\mathrm{W}_{0 \mathrm{AC}}=W_{Q A}+W_{A C}=125 \mathrm{~J}\)

Work done along path OBis

$$ \begin{aligned} \mathrm{W}_{\mathrm{OB}} &=\int F \cdot d x+\int F \cdot d y+\int F \cdot d z \\ &=0+0+0=0 \end{aligned} $$

Work done along path \(\mathrm{BC}_{1 \mathrm{~s}}\)

$$ \begin{aligned} \mathrm{W}_{\mathrm{BC}} &=\int_{0}^{5} F d x+\int_{5}^{5} F \cdot d y+\int_{0}^{0} F \cdot d z \\ &=\int_{0}^{5}(2 y) d x+0+0 \\ &=\int_{0}^{5}(10) \cdot d x \\ &=10 \times 5 \\ &=50 \mathrm{~J} \end{aligned} $$

The work done along OBC is \(W_{0 B C}=W_{O B}+W_{B C}=50 J\)

The work done along \(O C\) is

$$ \begin{aligned} W_{0 C} &=\int_{0}^{5} F \cdot d x+\int_{0}^{5} F \cdot d y+\int_{0}^{0} F \cdot d z \\ &=\int_{0}^{5}(2 y) \cdot d x+\int_{0}^{5}\left(x^{2}\right) \cdot d y+\int_{0}^{0} F \cdot d z \end{aligned} $$

Since \(y=x\)

$$ \begin{array}{l} =\int_{0}^{5}(2 x) d x+\int_{0}^{5}\left(y^{2}\right) \cdot d y+\int_{0}^{0} F \cdot d z \\ =\left(x^{2}\right)_{0}^{5}+\left(\frac{y^{3}}{3}\right)_{0}^{5} \\ =25+\frac{125}{3} \\ =66.67 \mathrm{~J} \end{array} $$