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Question

$$A$$ block of mass $$0.18$$ kg is attached to a spring of force-constant $$2\ N/m$$. The coefficient of friction between the block and the floor is $$0.1$$. Initially, the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of $$0.06 \ m$$ and comes to rest for the first time. The initial velocity of the block in $$m/s$$ is $$V=\dfrac{N}{10}$$. Then $$N$$ is:

28792.PNG


A
4
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B
5
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C
6
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D
7
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Solution

The correct option is A 4
applying work energy theorem
$$- \dfrac{1}{2} k x^{2}$$ $$-$$ $$ \mu mg x$$ $$=$$ $$\dfrac{1}{2} m v^{2}$$
 $$v=$$ $$\dfrac{4}{10}$$
N $$=$$ 4

Physics

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