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Standard X

Physics

Power

A 100 N for...

Question

A $100 N$ force acts horizontally on a block of mass $10 k g$ placed on a horizontal rough table of coefficient of friction $μ = 0.5$ . If $g$ at the place is $10 {m s}^{- 2}$ the acceleration of the block is:

Z e r o

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10 {m s}^{- 2}

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5 {m s}^{- 2}

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5.2 {m s}^{- 2}

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Solution

The correct option is C $5 {m s}^{- 2}$
limiting friction= $μ m g = 0.5 \times 10 g = 50 N$
net force = $100 - 50 = 50 N$
$50 = 10 a$
$a = 5 m / s^{2}$

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A 100 N force acts horizontally on a block of mass 10 kg placed on a horizontal rough table of coefficient of friction μ=0.5. If g at the place is 10ms−2 the acceleration of the block is:

The correct option is C 5 ms−2limiting friction=μmg=0.5×10g=50Nnet force =100−50=50N50=10aa=5m/s2

A $100 N$ force acts horizontally on a block of mass $10 k g$ placed on a horizontal rough table of coefficient of friction $μ = 0.5$ . If $g$ at the place is $10 {m s}^{- 2}$ the acceleration of the block is:

The correct option is C $5 {m s}^{- 2}$
limiting friction= $μ m g = 0.5 \times 10 g = 50 N$
net force = $100 - 50 = 50 N$
$50 = 10 a$
$a = 5 m / s^{2}$