SATHEE: Laws Of Motion Question 133

Laws Of Motion Question 133

Question: 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

Options:

A) $m g \cos θ$

B) $m g \sin θ$

C) $m g$

D) $m g / \cos θ$

Show Answer

Answer:

Correct Answer: D

Solution:

[d] When the whole system is accelerated towards left, then pseudo force (ma) works on a block towards right. For the condition of equilibrium: $m g \sin θ = m a \cos θ$

$\Rightarrow a = \frac{g \sin θ}{\cos θ}$ Therefore, force exerted by the wedge on the block $N = m g \cos θ + m a \sin θ$

$= m g \cos θ + m (\frac{g \sin θ}{\cos θ}) \sin θ$

$= \frac{m g (c o s^{2} θ + s i n^{2} θ)}{\cos θ} = \frac{m g}{\cos θ}$

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Laws Of Motion Question 133

Question: 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

Options:

Answer:

Solution:

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Question: 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