Question

A variable force, given by the two dimensional vector $\vec{F}=\left(3x^2\hat{i}+4\hat{j}\right)$, acts on a particle. The force is in newton and $x$ is in meter. 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 meter)

• A. -7 J
• B. Zero
• C. +7 J
• D. +19 J

Answer 1

Answer: (C)

Given, the two-dimensional force
$F=3x^2\hat{i}+4\hat{j}$
$\vec{r}=x\hat{i}+y\hat{j}$
$d\vec{r}=dx\hat{i}+dy\hat{j}$
Kinetic energy $=$ Work done
$W=\int F\cdot d\vec{r}$
$=\underset{(2,3)}{\overset{(3,0)}{\int }}\left(3x^2\hat{i}+4\hat{j}\right)\cdot (dx\hat{i}+dy\hat{j})$
$=\underset{(2,3)}{\overset{(3,0)}{\int }}3x^{2}dx+4dy$
$=\underset{2}{\overset{3}{\int }}3x^{2}dx+{\int }_3^{0}4dy$
$={\left[x^3\right]}_2^3+4[y{]}_3^0$
$=(27-8)+4(-3)$
$=19-12=7J$