Question

A block of mass $1 \,kg$ is attached to one end of a spring of force constant $k=20\, N / m$. The other end of the spring is attached to a fixed rigid support. This spring block system is made to oscillate on a rough horizontal surface $(\mu=0.04)$. The initial displacement of the block from the equilibrium position is $a=30$ $cm$. How many times the block passes from the mean position before coming to rest ? (Take $g=10 \,m / s ^{2}$ )

• A. 11
• B. 7
• C. 6
• D. 15

Answer 1

Answer: (B)

By energy conservation if block goes to a distance $x_{1}$ on other side then we use

$0+\frac{1}{2} k x^{2}-\frac{1}{2} k x_{1}^{2}-\mu m g\left(x+x_{1}\right)=0$
$\frac{1}{2}(20)(0.09)-\frac{1}{2}(20) x_{1}^{2}-(0.04)(1)(10)\left(0.3+x_{1}\right)=0$
$0.9-10 x_{1}^{2}-0.12-0.4 x_{1}=0$
$10 x_{1}^{2}+0.4 x_{1}-0.78=0$
$\Rightarrow x_{1}=0.26\, m$
Again if block covers $a$ distance $x_{2}$ to the right of mean position,
$0+\frac{1}{2} k x_{1}^{2}-\frac{1}{2} k x_{2}^{2}-\mu m g\left(x_{1}+x_{2}\right)=0$
$0+\frac{1}{2}(20)(0.26)^{2}-\frac{1}{2}(20) x_{2}^{2}-(0.04)(1)(10)\left(0.26+x_{2}\right)=0 $
$0.676-10 x_{2}^{2}-0.104-0.4 x_{2}=0$
$10 x_{2}^{2}+0.4 x_{2}-0.572=0$
$\Rightarrow x_{2}=0.22\, m$
The distance covered by block from mean position decreases each time by $0.04 \,m$.

Thus, block passes the mean position seven times before coming to rest when $k x < \mu m g$.