Question

A body of mass $1\text{kg}$ moving with a velocity of $5\text{m/s}$ hits a spring on its free end. The other end of spring is fixed and it has a force constant of $400\text{N/m}$. The body comes to rest after compressing the spring by '$x$' $\text{m}$, the value of '$x$' $\text{m}$ is

• A. $0.55\text{m}$
• B. $0.35\text{m}$
• C. $0.45\text{m}$
• D. $0.25\text{m}$

Answer 1

The correct option is D $0.25\text{m}$

Given $u=5\text{m/s}$
Force constant $k=400\text{N/m}$
let $x$ be compression in the spring when body comes to rest.
Applying law of conservation of energy,
$K_f-K_i=U_i-U_f$
($K_f=0$ as final velocity is zero ,$U_i=0$ as spring is not streched initially)
$-\frac{1}{2}m{u}^2=-\frac{1}{2}k{x}^2$
$\therefore x=u\sqrt{\frac{m}{k}}=5\sqrt{\frac{1}{400}}=\frac{5}{20}=0.25\text{m}$