A particle undergoes from position O(0,0,0) to via path in plane under the action of a force which varies

Question

A particle undergoes from position O(0,0,0) to $A (a, 2 a, 0)$ via path $y = \frac{2 x^{2}}{a}$ in $x - y$ plane under the action of a force which varies with particle's $(x, y, z)$ coordinate as $\vec{F} = x^{2} y \hat{i} + y z^{2} e^{2 z} \hat{j} - (\frac{z}{x + 2 y}) \hat{k} .$ Work done by the force $\vec{F}$ is: (all symbols have their usual meaning and they are in SI unit.)

Smart Achievers · Accepted Answer

(4)

$\vec{F} = x^{2} y \hat{i} + y z^{2} e^{2 z} \hat{j} - (\frac{z}{x + 2 y}) \hat{k}$

$d \vec{r} = d x \hat{i} + d y \hat{j} + d z \hat{k}$

$d w = \vec{F} \cdot d \vec{r} = x^{2} y d x + y z^{2} e^{2 z} d y - (\frac{z}{x + 2 y}) d z$

$for the given path z = 0, y = \frac{2 x^{2}}{a}$

$d w = x^{2} y d x = \frac{2 x^{4}}{a} d x$

$W = \int d w = \frac{2}{a} \int_{0}^{a} x^{4} d x = \frac{2 a^{4}}{5}$

Solution

1 $\frac{4 a^{4}}{5}$

2 $\frac{a^{4}}{5}$

3 $\frac{a^{3}}{4}$

4 $\frac{2 a^{4}}{5}$

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Solution

Q. A particle undergoes from position O(0,0,0) to A(a,2a,0) via path y=2x2a in x-y plane under the action of a force which varies with particle's (x,y,z) coordinate as F→=x2yi^+yz2e2zj^-(zx+2y)k^. Work done by the force F→ is: (all symbols have their usual meaning and they are in SI unit.)

1 4a45

2 a45

3 a34

4 2a45

Ans. (4) F→=x2yi^+yz2e2zj^-(zx+2y)k^ dr→=dxi^+dyj^+dzk^ dw=F→·dr→=x2y dx+yz2e2z dy-(zx+2y)dz for the given path z=0, y=2x2a dw=x2y dx=2x4a dx W=∫dw=2a∫0ax4 dx=2a45

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1 $\frac{4 a^{4}}{5}$

2 $\frac{a^{4}}{5}$

3 $\frac{a^{3}}{4}$

4 $\frac{2 a^{4}}{5}$