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Standard XII

Physics

Problem Solving

A wooden bloc...

Question

A wooden block of mass 20kg slides without acceleration down a rough inclined plane which makes angle of $20^{0}$ with the horizontal. Find the acceleration of the block when the inclination of a plane is increased to $30^{0}$ .

1.81 m s^{- 2}

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3.62 m s^{- 2}

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0.90 m s^{- 2}

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0.45 m s^{- 2}

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Solution

The correct option is A $1.81 m s^{- 2}$
For the first case the acceleration is $0$ , so $μ = t a n 20^{o}$
When, $θ = 30^{o}, a = g s i n 30^{o} - g c o s 30^{o} t a n 20^{o} = 9.8 \times 0.5 - 9.8 \times \frac{\sqrt{3}}{2} \times 0.36 = 1.81 m s^{- 2}$

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Similar questions

A 10 kg block slides, without acceleration, down a rough inclined plane making an angle of $20^{\circ}$ with the horizontal. Find the acceleration when the angle of inclination is increased to $30^{\circ}$ (g = 9.8 $m s^{- 2}$ )

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A wooden block of mass 20kg slides without acceleration down a rough inclined plane which makes angle of 200 with the horizontal. Find the acceleration of the block when the inclination of a plane is increased to 300.

The correct option is A 1.81ms−2For the first case the acceleration is 0, so μ=tan20oWhen, θ=30o,a=gsin30o−gcos30otan20o=9.8×0.5−9.8×√32×0.36=1.81 ms−2

A wooden block of mass 20kg slides without acceleration down a rough inclined plane which makes angle of $20^{0}$ with the horizontal. Find the acceleration of the block when the inclination of a plane is increased to $30^{0}$ .

The correct option is A $1.81 m s^{- 2}$
For the first case the acceleration is $0$ , so $μ = t a n 20^{o}$
When, $θ = 30^{o}, a = g s i n 30^{o} - g c o s 30^{o} t a n 20^{o} = 9.8 \times 0.5 - 9.8 \times \frac{\sqrt{3}}{2} \times 0.36 = 1.81 m s^{- 2}$