A block is lying on an inclined plane which makes an angle of $60^{o}$ with the horizontal. If coefficient of friction between block and plane is 0.25 and $g = 10 m s^{-} 2$ , then the acceleration of block when it moves along the plane will be:

2.50 m s^{- 2}

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5.00 m s^{- 2}

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7.25 m s^{- 2}

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8.66 m s^{- 2}

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Solution

The correct option is C $7.25 m s^{- 2}$
From the above figure

Let, m be the mass and a is the acceleration produced. The frictional force also acts. The net force on the block is:

$F_{n e t} = m a = m g sin θ - μ m g cos θ$

$a = g (sin θ - μ cos θ)$

$= 10 (sin 60 - 0.25 \times cos 60)$

$= 10 (0.85 - 0.125)$

$= 7.25 m / s^{2}$

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A block is lying on an inclined plane which makes an angle of 60o with the horizontal. If coefficient of friction between block and plane is 0.25 and g=10ms−2, then the acceleration of block when it moves along the plane will be:

The correct option is C 7.25 ms−2From the above figure Let, m be the mass and a is the acceleration produced. The frictional force also acts. The net force on the block is: Fnet=ma=mgsinθ−μmgcosθ a=g(sinθ−μcosθ) =10(sin60−0.25×cos60) =10(0.85−0.125) =7.25m/s2

A block is lying on an inclined plane which makes an angle of $60^{o}$ with the horizontal. If coefficient of friction between block and plane is 0.25 and $g = 10 m s^{-} 2$ , then the acceleration of block when it moves along the plane will be: