Question

A force acting on a particle moving in the $x - y$ plane is given by $\vec{F} = (2 y \hat{i} + x^{2} \hat{j}) N,$ where $x$ and $y$ are in meters. The particle moves from the origin to a final position having coordinates $x = 5.00 m$ and $y = 5.00 m$ as shown in the figure. Calculate the work done by $\vec{F}$ on the particle as it moves along

(a) $O A C$

(b) $O B C$

(c) $O C$

Is $\vec{F}$ conservative or non-conservative? Explain.

Question Image

(a) $O A C$

(b) $O B C$

(c) $O C$

Is $\vec{F}$ conservative or non-conservative? Explain.

Accepted Answer

(a) Given that

Applied force $\vec{F} = (2 y \hat{i} + x^{2} \hat{j}) N,$

For each of the paths from $O$ to $C,$ the work done is given by

$W = \int \vec{F} \cdot \vec{d s}$

Here $\vec{F} = (2 y \hat{i} + x^{2} \hat{j})$ and $\vec{d s} = (d x \hat{i} + d y \hat{j})$

Hence $W = \int (2 y \hat{i} + x^{2} \hat{j}) (d x \hat{i} + d y \hat{j})$

$\Rightarrow W = \int_{0}^{5} 2 y d x + \int_{0}^{5} x^{2} d y . . . . (i)$

The path $O A C$ can be break up into two paths, from $O$ to $A$ and then from $A$ to $C .$

First considering path $O$ to $A,$ in this path $y = 0 .$ Hence, from equation $(i)$ it is clear $W_{OA} = 0$

Now consider path $A$ to $C$ in this path $x$ is constant (i.e., $d x = 0$ ) and is equal to $x = 5 m .$ From equation $(i)$

$W_{AC} = x^{2} \int_{0}^{5} d y = {(5)}^{2} {[y]}_{0}^{5} = 125 J$

Hence, total work done $W_{OAC} = 0 + 125 = 125 J$

(b) Now consider the path $O B C,$ this path also break up into two paths $O$ to $B$ and then $B$ to $C .$ Considering path $O$ to $B,$ in this path $x = 0 .$ Hence, from equation (i) it is clear $W_{O B} = 0$ . Now considering path $B$ to $C,$ in this path $y$ is constant (i.e. $d y = 0$ ) from equation (i)

$W_{B C} = 2 y \int_{0}^{5} d x = 2 \times 5 \times {[x]}_{0}^{5} = 50 J$

Hence total work done in this path $W_{O B C} = 0 + 50 = 50 J$

(c) Motion of the particle from $O$ to $C,$ in this path both $x$ and $y$ are changing. In this path $x = y \Rightarrow d x = d y$ From equation (i);

$W_{O C} = \int_{0}^{5} (2 x + x^{2}) d x = 2 {[\frac{x^{2}}{2}]}_{0}^{5} + {[\frac{x^{3}}{3}]}_{0}^{5}$

$= 25 + \frac{1}{3} \times 125 = \frac{200}{3} J$

B M Sharma Solutions for Chapter: Work, Energy and Power, Exercise 37: ILLUSTRATION 8.35

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Work, Energy and Power, Exercise 37: ILLUSTRATION 8.35

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