A horizontal force just sufficient to move a body of mass $4 k g$ lying on a rough horizontal surface, is applied on it . Coefficients of static and kinetic friction are $0.8$ and $0.6$ respectively. If the force continues to act even after the body has started moving, the acceleration of the body is:
$(g = 10 m s^{- 2})$

6 m s^{- 2}

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8 m s^{- 2}

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2 m s^{- 2}

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4 m s^{- 2}

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Solution

The correct option is A $2 m s^{- 2}$
Force responsible for increase in velocity $= (μ_{s} - μ_{k}) m g$
$= (0.8 - 0.6) \times 4 \times 10$
$= 0.2 \times 4 \times 10 = 8 N$

Therefore, $a = \frac{8}{4} = 2 m s^{- 2}$

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A block of mass 5 kg is lying on a rough horizontal surface. The coefficients of static and kinetic friction between the block and the surface respectively are 0.7 and 0.5. A horizontal force just sufficient to move the block is applied to it. If the force continues to act even after the block has started moving, what will be the acceleration of the block? Take $g = 10 m s^{- 2}$ .

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)

A body of mass $2$ kg is placed on a horizontal surface having kinetic friction $0.4$ and static friction $0.5$ . If the horizontal force applied on the body is $2.5$ N, the frictional force acting on the body will be $(g = 10 m / s^{2})$

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The correct option is A 2ms−2Force responsible for increase in velocity =(μs−μk)mg =(0.8−0.6)×4×10 =0.2×4×10=8 N Therefore, a=84=2 ms−2

The correct option is A $2 m s^{- 2}$
Force responsible for increase in velocity $= (μ_{s} - μ_{k}) m g$
$= (0.8 - 0.6) \times 4 \times 10$
$= 0.2 \times 4 \times 10 = 8 N$

Therefore, $a = \frac{8}{4} = 2 m s^{- 2}$