Question #11eb9

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Question #11eb9

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Answer 1 · 2017-03-25T06:42:46

Given
drawn drawn
$m \to mass of the sliding block = 2 k g$ $m->"mass of the sliding block"=2kg$

$v \to speed of the sliding block = 4 m / s$ $v->"speed of the sliding block"=4m"/"s$

$F_{friction} \to force of kinetic friction = 15 N$ $F_"friction"->"force of kinetic friction"=15N$

$k \to spring constant of the spring = 10^{4} N / m$ $k->"spring constant of the spring"=10^4N"/"m$

If the given speed be at the moment of colliding the block on the spring and it moves $x$ $x$ m before coming to rest, then by considering the conservation of energy we can say that total initial KE of the block will be equal to the sum of gain in PE of the spring due to compression by $x$ $x$ m and work done against the force of friction due its displacement $x$ $x$ m on the floor.

So

$\frac{1}{2} m v^{2} = F_{friction} \times x + \frac{1}{2} k x^{2}$ $1/2mv^2=F_"friction"xx x+1/2kx^2$

$\Rightarrow k x^{2} + 2 \times F_{friction} \times x = m v^{2}$ $=>kx^2+2xxF_"friction"xx x=mv^2$

$\Rightarrow 10^{4} x^{2} + 2 \times 15 \times x = 2 \times 4^{2}$ $=>10^4x^2+2xx15xx x=2xx4^2$

$\Rightarrow {(100 x)}^{2} + 2 \times (100 x) \times (0.15) + {(0.15)}^{2} = 2 \times 4^{2} + ({0.15}^{2}$ $=>(100x)^2+2xx(100x)xx (0.15)+(0.15)^2=2xx4^2+(0.15^2$

$\Rightarrow {(100 x + 0.15)}^{2} = 2 \times 4^{2} + ({0.15}^{2}) = 32.0225$ $=>(100x+0.15)^2=2xx4^2+(0.15^2)=32.0225$

$\Rightarrow (100 x + 0.15) = \sqrt{32.0225}$ $=>(100x+0.15)=sqrt32.0225$

$\Rightarrow x = \frac{\sqrt{32.0225} - 0.15}{100} m = (\sqrt{32.0225} - 0.15) c m$ $=>x=(sqrt32.0225-0.15)/100m =(sqrt32.0225-0.15)cm$

$\Rightarrow x = 5.5 c m$ $=>x=5.5cm$ , Option (2)

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Question #11eb9

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