The work done by the force $F = A (y^{2} \hat{i} + 2 x^{2} \hat{j})$ , where $A$ is a constant and $x$ & $y$ are in meters around the path shown is :

Question

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Answer 1

Variable force is apllied across the path $O A$ to $A B$ to $B C$ to $C O$

$W = \int \overset{ˉ}{F} . d \overset{r}{ˉ}$

$= \int A (y^{2} \overset{ˉ}{i} + 2 x^{2} \overset{ˉ}{j}) . (d x \overset{ˉ}{i} + d y . \overset{ˉ}{j})$

$= \int A (y^{2} d x + 2 x^{2} d y)$

Now following the path of displacement along $O A$ the $y = 0$

$W_{O A} = \int_{x = 0}^{x = d} A (0 + 2 x^{2} . 0) = 0 + 0 = 0$

Similerly, for $W_{A B}, x = d$

$W_{A B} = A [0 + 2 d^{2} d] = 2 A d^{3}$

Similarly, for $W_{B C}, x = d, y = d$

$W_{B C} = \int_{x = d}^{x = 0} A (d^{2} d x + 0)$

$W_{B C} = A [d^{2} (- d) + 0] = - A d^{3}$

$W_{C O} = A [0 + 0]$

$W = 0 + 2 A d^{3} - A d^{3} + 0 = A d^{3}$

Answer 2

Variable force is apllied across the path

O A

to

A B

to

B C

to

C O

W = \int \overset{ˉ}{F} . d \overset{r}{ˉ}

= \int A (y^{2} \overset{ˉ}{i} + 2 x^{2} \overset{ˉ}{j}) . (d x \overset{ˉ}{i} + d y . \overset{ˉ}{j})

= \int A (y^{2} d x + 2 x^{2} d y)

Now following the path of displacement along

O A

the

y = 0

W_{O A} = \int_{x = 0}^{x = d} A (0 + 2 x^{2} . 0) = 0 + 0 = 0

Similerly, for

W_{A B}, x = d

W_{A B} = A [0 + 2 d^{2} d] = 2 A d^{3}

Similarly, for

W_{B C}, x = d, y = d

W_{B C} = \int_{x = d}^{x = 0} A (d^{2} d x + 0)

W_{B C} = A [d^{2} (- d) + 0] = - A d^{3}

W_{C O} = A [0 + 0]

W = 0 + 2 A d^{3} - A d^{3} + 0 = A d^{3}