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5.Work, Energy, Power and Collision
hard
An object of mass ' $m$ ' initially at rest on a smooth horizontal plane starts moving under the action of force $F=2 N$. In the process of its linear motion, the angle $\theta$ (as shown in figure) between the direction of force and horizontal varies as $\theta= kx$, where $k$ is a constant and $x$ is the distance covered by the object from its initial position. The expression of kinetic energy of the object will be $E =\frac{ n }{ k } \sin \theta$. The value of $n$ is $.....$
A
$1$
B
$3$
C
$4$
D
$2$
(JEE MAIN-2023)
Solution
$F \cos \theta= ma$
$2 \cos ( kx )=\frac{ mvdv }{ dx }$
$\int \limits_0^v v d v=2 \int \limits_0^x \cos (k x) d x$
$K . E .=\frac{2}{ k } \sin \theta$
$n=2$
Standard 11
Physics
