Question

A block of mass $0.18kg$ is attached to a spring of force constant $2N{m}^{-1}$. 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 Fig. The block slides a distance of $0.06m$ and comes to rest for the first time. The initial velocity of the block in $m{s}^{-1}$ is V = N/10. Then N is :

• A. 4
• B. 5
• C. 6
• D. 7

Answer 1

Answer: (A)

4

Given

Mass of block, $m=0.18kg$

Spring Constant, $k=2N/m$

coefficient of friction, $\mu =0.1$

Distance travelled $x=0.06m$

Initial velocity, $v=\frac{N}{10}m/s$

To Find

N

Procedure

The forces acting on the spring,

during its motion are, weight mg,

normal force N (=mg) and friction,

$f(=\mu N=\mu mg).$

$\therefore$ Work done by friction $=fx=\mu mgx...\left(1\right)$

Potential Energy in the spring $=\frac{1}{2}k{x}^2\to \left(2\right)$

Initial kinetic energy $=\frac{1}{2}m{v}^2\to \left(3\right)$

The initial kinetic energy is stored as

potential energy in the spring, and as

energy lost against friction.

$\therefore \frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2+\mu mgx$

$\Rightarrow V=\sqrt{\frac{k{x}^2}{m}+2\mu gx}$

$=\sqrt{\frac{2\times {0.06}^2}{0.18}+2\times 0.1\times 10\times 0.06}$

$\Rightarrow v=0.4$

Since, $v=\frac{N}{10}=0.4,$ we see $N=4$