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Standard XII

Physics

Range

A block of ma...

Question

A block of mass $1 k g$ is projected from the lowest point up along the inclined plane. If $g = 10 m s^{- 2}$ , the retardation experienced by the block is

\frac{15}{\sqrt{2}} m s^{- 2}

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\frac{5}{\sqrt{2}} m s^{- 2}

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\frac{10}{\sqrt{2}} m s^{- 2}

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Solution

The correct option is A $\frac{15}{\sqrt{2}} m s^{- 2}$
$N = m g c o s θ f = μ m g c o s θ a_{r} = \frac{m g s i n θ + μ m g c o s θ}{m} a_{r} = g (s i n θ + μ c o s θ) g (\frac{1}{\sqrt{2}} + \frac{0.5}{\sqrt{2}}) = \frac{10}{\sqrt{2}} + \frac{5}{\sqrt{2}} = \frac{15}{\sqrt{2}}$

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A block of mass 1 kg is projected from the lowest point up along the inclined plane. If g=10 ms−2, the retardation experienced by the block is

The correct option is A 15√2ms−2N=mgcosθf=μmgcosθar=mgsinθ+μmgcosθmar=g(sinθ+μcosθ)g(1√2+0.5√2)=10√2+5√2=15√2

A block of mass $1 k g$ is projected from the lowest point up along the inclined plane. If $g = 10 m s^{- 2}$ , the retardation experienced by the block is

The correct option is A $\frac{15}{\sqrt{2}} m s^{- 2}$
$N = m g c o s θ f = μ m g c o s θ a_{r} = \frac{m g s i n θ + μ m g c o s θ}{m} a_{r} = g (s i n θ + μ c o s θ) g (\frac{1}{\sqrt{2}} + \frac{0.5}{\sqrt{2}}) = \frac{10}{\sqrt{2}} + \frac{5}{\sqrt{2}} = \frac{15}{\sqrt{2}}$