Question

A variable force, given by the two dimensional vector $\vec{ F }=\left(3 x^{2} \hat{ i }+4 \hat{ j }\right)$, 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
$F =3 x^{2} \hat{ i }+4 \hat{ j }$
$\vec{ r }=x \hat{ i }+y \hat{ j }$
$d \vec{ r }=d x \hat{ i }+d y \hat{ j }$
Kinetic energy $=$ Work done
$W=\int F \cdot d \vec{ r }$
$=\int\limits_{(2,3)}^{(3,0)}\left(3 x^{2} \hat{ i }+4 \hat{ j }\right) \cdot(d x \hat{ i }+d y \hat{ j })$
$=\int\limits_{(2,3)}^{(3,0)} 3 x^{2} d x+4 d y$
$=\int\limits_{2}^{3} 3 x^{2} d x+\int_{3}^{0} 4 d y$
$=\left[x^{3}\right]_{2}^{3}+4[y]_{3}^{0}$
$=(27-8)+4(-3)$
$=19-12 =7\, J$