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

Work Power and Energy PYQ — JEE Advanced 2011 · Shift 2 · Q49 with Solution

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Correct answer: 4

We need the initial velocity $V$ .

Initially, the block has kinetic energy

K_i = \frac12 mV^2

As the block moves a distance $x$ , this kinetic energy is spent in:

At the first stop, final kinetic energy is zero, so

\frac12 mV^2 = \frac12 kx^2 + f x

where friction force

f = \mu mg

So,

\frac12 mV^2 = \frac12 kx^2 + \mu mgx

First, spring potential energy:

\frac12 kx^2 = \frac12 \cdot 2 \cdot (0.06)^2 = (0.06)^2 = 0.0036\,\text{J}

Now friction work:

\mu mgx = 0.1 \cdot 0.18 \cdot 10 \cdot 0.06

Using $g = 10\,\text{m/s}^2$ ,

\mu mgx = 0.0108\,\text{J}

Hence,

\frac12 mV^2 = 0.0036 + 0.0108 = 0.0144

Now substitute $m=0.18$ :

\frac12 \cdot 0.18\,V^2 = 0.0144

0.09V^2 = 0.0144

V^2 = \frac{0.0144}{0.09} = 0.16

V = 0.4\,\text{m/s}

Given

V = \frac{N}{10}

So,

0.4 = \frac{N}{10}

N = 4

\boxed{N=4}

The derived answer matches the stored correct answer.