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  4. A 100 N force acts horizontally on a block of 10 kg placed on a horizontal rough surface of coefficient of friction μ= 0.5. If the acceleration due to gravity ( g ) is taken as 10 ms - 2 , the acceleration of the block (in ms -2 ) is

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Q. A $100 \,N$ force acts horizontally on a block of $10 \,kg$ placed on a horizontal rough surface of coefficient of friction $\mu=$ $0.5$. If the acceleration due to gravity $( g )$ is taken as $10 \,ms ^{-}$ 2 , the acceleration of the block (in $ms ^{-2}$ ) is

Laws of Motion

Solution:

$a =\frac{ F -\mu R }{ m }=\frac{100-0.5 \times(10 \times 10)}{10}=5\, ms ^{-2}$