Question 254: Work, Energy and Power & Circular Motion

Question

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 $$	heta$$ (as shown in figure) between the direction of force and horizontal varies as $$	heta=\mathrm{k}x$$, 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 $$E = {n \over k}\sin 	heta $$. The value of n is ___________.

Accepted Answer

$$
	ext { Work done }=\Delta \mathrm{K} . \mathrm{E}
$$
$$
\begin{aligned}
& 	herefore \int F \cdot d x=\frac{1}{2} m v^{2}=E \\
& 	herefore E=\int_{0}^{x} 2 \cos (k x) d x \\
& E=\frac{2}{k}[\sin k x]_{0}^{x} \\
& =\frac{2}{k} \sin k x \\
& =\frac{2 \sin 	heta}{k}
\end{aligned}
$$

Work, Energy and Power & Circular Motion

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