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Physics Question on laws of motion

A block of mass mm is placed on a smooth wedge of inclination θ\theta. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block will be (gg is acceleration due to gravity)

A

mgcos⁡θ;mg \cos \theta;

B

mgsin⁡θ;mg \sin\theta;

C

mg

D

mgcos θ\frac {mg} {cos\, \theta}

Answer

mgcos θ\frac {mg} {cos\, \theta}

Explanation

Solution

The wedge is given an acceleration to the left.
∴\therefore The block has a pseudo acceleration to the right, pressing against the wedge because of which the block is not moving.

∴mg sin⁡ θ=ma cos⁡ θ\therefore mg\,\sin\,\theta=ma \,\cos\,\theta or a=g sin⁡ θcos⁡ θa=\frac{g\,\sin\,\theta}{\cos\,\theta}
Total reaction of the wedge on the block is
N=mg cos⁡ θ+ma sin⁡ θN=mg \,\cos\,\theta+ma \,\sin\,\theta.
or N=mg cos⁡ θ+mg sin⁡ θ⋅sin⁡ θcos⁡ θN=mg\,\cos\, \theta+\frac{mg\,\sin\,\theta \cdot \sin\,\theta}{\cos\,\theta}.
N=mg(cos⁡2 θ+sin⁡2 θ)cos⁡ θ=mgcos⁡ θN=\frac{mg(\cos^2 \,\theta+\sin^2\,\theta)}{\cos\,\theta}=\frac{mg}{\cos\,\theta}