Work Power and Energy 3 Question 5

5. 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 $v=\frac{N}{10}$. Then $N$ is

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(2011)

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Answer:

Correct Answer: 5. (4)

Solution:

  1. Decrease in mechanical energy $=$ Work done against friction

$\therefore \frac{1}{2} m v^{2}-\frac{1}{2} k x^{2}=\mu m g x$ or $v=\sqrt{\frac{2 \mu m g x+k x^{2}}{m}}$

Substituting the values, we get

$$ v=0.4 m / s=\frac{4}{10} m / s \Rightarrow N=4 $$



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