$$AB$$ is a quarter of a smooth circular track of radius $$R=2m$$ as shown in figure. A particle $$P$$ of mass $$m=5kg$$ moves along the track from $$A\ to\ B$$ under the action of a force which always directed toward point $$B$$ and has magnitude $$10\ N$$. Find the work done by force in moving the object from $$A\ to\ B$$.

The force has constant magnitude $$F=10\,N$$ and is always directed towards B.

Let the position of P on the quarter circle be represented by the angle $$\theta$$, measured from OB. Then

$$ds=R\,d\theta$$

The angle between the tangent at P and the force is $$\frac{\theta}{2}$$. Therefore, the work done is

$$W=\int_A^B F\cos\frac{\theta}{2}\,ds$$

Here, $$R=2\,m$$ and $$\theta$$ varies from $$\frac{\pi}{2}$$ to $$0$$.

$$W=FR\int_0^{\pi/2}\cos\frac{\theta}{2}\,d\theta$$

$$W=FR\left[2\sin\frac{\theta}{2}\right]_0^{\pi/2}$$

$$W=10\times2\times2\sin\frac{\pi}{4}$$

$$W=20\sqrt{2}\,J$$

Therefore,  $${W=20\sqrt{2}\,J}$$

Hence, the correct option is D.