The retarding acceleration of $4 m / s^{2}$ due to frictional force stops a block of mass $40 k g$ moving on a rough horizontal surface. The coefficient of friction between the block and the surface is

0.3

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0.4

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0.55

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0.65

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Solution

The correct option is B $0.4$
As we know, coefficient of friction $μ = \frac{f}{N}$
where $f$ is the force of friction acting on the block.

$f = m a$ for retardation $a$

$\Rightarrow μ = \frac{m a}{m g} = \frac{a}{g}$
$a = 4 m s^{- 2} (given)$
$∴ μ = \frac{4}{10} = 0.4$

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The retarding acceleration of 4 m/s2 due to frictional force stops a block of mass 40 kg moving on a rough horizontal surface. The coefficient of friction between the block and the surface is

The correct option is B 0.4As we know, coefficient of friction μ=fN where f is the force of friction acting on the block. f=ma for retardation a ⇒ μ=mamg=ag a=4 ms−2(given) ∴ μ=410=0.4

The retarding acceleration of $4 m / s^{2}$ due to frictional force stops a block of mass $40 k g$ moving on a rough horizontal surface. The coefficient of friction between the block and the surface is

The correct option is B $0.4$
As we know, coefficient of friction $μ = \frac{f}{N}$
where $f$ is the force of friction acting on the block.

$f = m a$ for retardation $a$

$\Rightarrow μ = \frac{m a}{m g} = \frac{a}{g}$
$a = 4 m s^{- 2} (given)$
$∴ μ = \frac{4}{10} = 0.4$