SATHEE: Work Power and Energy 3 Question 5

Work Power and Energy 3 Question 5

5. A block of mass $0.18 k g$ 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:

Decrease in mechanical energy $=$ Work done against friction

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

Substituting the values, we get

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

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Work Power and Energy 3 Question 5

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