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Q.

A block of mass m is placed on a smooth wedge of inclination $θ$ . 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 (g is acceleration due to gravity) will be

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a

$m g / \cos θ$

b

$m g \cos θ$

c

$m g \sin θ$

d

$m g$

answer is A.

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Detailed Solution

$N = m a \sin θ + m g \cos θ$

Question Image

Also $m g \sin θ = m a \cos θ$

From (2) $a = g \tan θ$

$∴ N = m g \frac{\sin^{2} θ}{\cos θ} + m g \cos θ$ ,

$o r N = \frac{m g}{\cos θ}$

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