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  4. 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 un-stretched. 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 = N / 10 . Then N is

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Q. 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 un-stretched. 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 $V = N / 10 .$ Then $N$ isPhysics Question Image

JEE AdvancedJEE Advanced 2011

Solution:

Applying work energy theorem
$-\frac{1}{2} kx ^{2}-\mu mg x =-\frac{1}{2} m V ^{2}$
$\Rightarrow V =\frac{4}{10}$
$N =4$