Work Power Energy Question 3
Question 3 - 25 January - Shift 1
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
Smooth horizontal surface
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Answer: (2)
Solution:
Formula: Kinetic Energy
$F \cos \theta=ma$
$2 \cos (kx)=\frac{mvdv}{dx}$
$\int_0^{v} v d v=2 \int_0^{x} \cos (k x) d x$
$\frac{mv^{2}}{2}=\frac{2}{k} \sin kx$
K.E. $=\frac{2}{k} \sin \theta$
$\mathbf{n}=\mathbf{2}$
