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Byju's Answer

Standard XII

Physics

Constrained Motion : Intuitive Approach

A block of ma...

Question

A block of mass $m$ is pulled by a string acts with passes over a smooth pulley $P$ fitted with the wedge of mass $M$ . A horizontal force $F = m g$ acts on the string. If the string is massless and frictionless, assuming frictionless contacting surfaces and $θ = cos - 1 4 / 5$ , then :

N = 4 / 5 m g

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N =

Normal reaction on block

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a_{1} = 17 g / 25

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a_{1} =

acceleration of

^{'} m^{'}

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Solution

The correct option is A $N = 4 / 5 m g$
Given $cos θ = \frac{4}{5}$
$m a = F - m g sin θ$
$m a = m g (1 - \frac{3}{5})$
$a = g (\frac{2}{5})$
and $N = m g cos θ = m g (\frac{4}{5})$
Hence, the option $A$ is the correct answer.

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A block of mass m is pulled by a string acts with passes over a smooth pulley P fitted with the wedge of mass M. A horizontal force F=mg acts on the string. If the string is massless and frictionless, assuming frictionless contacting surfaces and θ=cos−1 4/5, then :

The correct option is A N=4/5 mgGiven cosθ=45ma=F−mgsinθma=mg(1−35)a=g(25)and N=mgcosθ=mg(45)Hence, the option A is the correct answer.

A block of mass $m$ is pulled by a string acts with passes over a smooth pulley $P$ fitted with the wedge of mass $M$ . A horizontal force $F = m g$ acts on the string. If the string is massless and frictionless, assuming frictionless contacting surfaces and $θ = cos - 1 4 / 5$ , then :

The correct option is A $N = 4 / 5 m g$
Given $cos θ = \frac{4}{5}$
$m a = F - m g sin θ$
$m a = m g (1 - \frac{3}{5})$
$a = g (\frac{2}{5})$
and $N = m g cos θ = m g (\frac{4}{5})$
Hence, the option $A$ is the correct answer.