JEE MAIN - Physics (2023 - 25th January Morning Shift - No. 19)
An object of mass 'm' initially at rest on a smooth horizontal plane starts moving under the action of force F = 2N. In the process of its linear motion, the angle $$\theta$$ (as shown in figure) between the direction of force and horizontal varies as $$\theta=\mathrm{k}x$$, 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 $$E = {n \over k}\sin \theta $$. The value of n is ___________.
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Answer
2
Explanation
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$$ \begin{aligned} & \therefore \int F \cdot d x=\frac{1}{2} m v^{2}=E \\\\ & \therefore E=\int_{0}^{x} 2 \cos (k x) d x \\\\ & E=\frac{2}{k}[\sin k x]_{0}^{x} \\\\ & =\frac{2}{k} \sin k x \\\\ & =\frac{2 \sin \theta}{k} \end{aligned} $$
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