A variable force, given by the 2-dimensional vector $\overrightarrow{F} = (3x^{2}\widehat{i} + 4\ \widehat{j})$, acts on a particle. The force is in newton and $x$ is in metre. What is the change in the kinetic energy of the particle as it moves from the point with coordinates (2,3) to (3,0) (The coordinates are in metres)

Question

Dr. Abubacker · Accepted Answer

(c) $\overrightarrow{F} = 3x^{2}\widehat{i} + 4\widehat{j},\overrightarrow{\ r} = x\widehat{i} + y\widehat{j}$
$	herefore d\overrightarrow{r} = dx\widehat{i} + dy\widehat{j}$
Work done, $W = \int_{}^{}{\overrightarrow{F}.d\overrightarrow{r} = \int_{(2,3)}^{(3,0)}{\left( 3x^{2}\widehat{i} + 4\widehat{j} ight).(dx\widehat{i} + dy\widehat{j})}}$
$= \int_{(2,3)}^{(3,0)}{\left( 3x^{2}dx + 4dy ight) = \left\lbrack x^{3} + 4y ightbrack_{(2,3)}^{(3,0)}}\ $
$= 3^{3} + 4 	imes 0 - (2^{3} + 4 	imes 3)$
$= 27 + 0 - (8 + 12) = 27 - 20 = + 7J$
According to work energy theorem
Change in the kinetic energy = Work done, $W = + 7J$

Explanation

Reviewed by Dr. Abubacker

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