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A block of mass 5kg is (i) pushed in case (A) and (ii) pulled in case (B), by a force F=20N, making and angle of $30^{0}$ with the horizontal, as shown in the figure. The coefficient of friction between the block, the floor is $μ = 0.2.$ The difference between the acceleration of the block, in case (B) and case (A) will be (Take, $g = 10 m s^{- 2}$ )

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3.2 $m s^{- 2}$

0.4 $m s^{- 2}$

0.8 $m s^{- 2}$

0 $m s^{- 2}$

answer is C.

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Detailed Solution

Case: 1

$N = m g + F \sin 30 ° = (5 \times 10) + (20 \times \frac{1}{2}) = 60 N$

$Force of friction, f = μ N = 0.2 \times 60 = 12 N [∵ μ = 0.2]$

$F_{net} = m a_{1} = Fcos 30 ° - f$

$a_{1} = \frac{10 \sqrt{3} - 12}{5} = 1 m s^{- 2}$ ,

Case 2 :

$\Rightarrow F_{net} = F \cos 30 ° - μ (mg - Fsin 30 °)$

$a_{2} = \frac{F_{n e t}}{m} = \frac{F \cos 30 ° - μ (m g - F \sin 30 °)}{m}$

$a_{2} = 1.8 m s^{- 2}$

So, difference $= a_{2} - a_{1} = 1.8 - 1 = 0.8 m s^{- 2}$

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