JEE MAIN - Physics (2022 - 27th July Evening Shift - No. 3)
A block of mass M slides down on a rough inclined plane with constant velocity. The angle made by the incline plane with horizontal is $$\theta$$. The magnitude of the contact force will be :
Mg
$$\mathrm{Mg} \cos \theta$$
$$\sqrt{\mathrm{Mg} \sin \theta+\mathrm{Mg} \cos \theta}$$
$$\operatorname{Mg} \sin \theta \sqrt{1+\mu}$$
Explanation
As the body is moving with constant velocity so forces acting on the body must be balanced.
$$\Rightarrow$$ Contact force from incline should balance weight of the body.
$$\Rightarrow$$ | Fcontact | = Mg
Comments (0)
