Two block of masses $m_1$ and $m_2$ connected with the help of a spring of spring constant $k$ initially to natural length as shown. A sharp impulse is given to mass $m_2$ so that it acquires a velocity $v_0$ towards right. If the system is kept an smooth floor then find the maximum elongation that the spring will suffer

- A
${v_0}\sqrt {\frac{{{m_1}{m_2}}}{{k({m_1} + {m_2})}}} $

- B
${v_0}\sqrt {\frac{{({m_1} + {m_2})}}{{k{m_1}{m_2}}}} $

- C
${v_0}\sqrt {\frac{{2{m_1}{m_2}}}{{k({m_1} + {m_2})}}} $

- D
$2{v_0}\sqrt {\frac{{{m_1}{m_2}}}{{k({m_1} + {m_2})}}} $

$A$ block of mass $m$ moving with a velocity $v_0$ on a smooth horizontal surface strikes and compresses a spring of stiffness $k$ till mass comes to rest as shown in the figure. This phenomenon is observed by two observers:

$A$: standing on the horizontal surface

$B$: standing on the block

According to observer $B$, the potential energy of the spring increases

A block of mass $m$ starts at rest at height $h$ on a frictionless inclined plane. The block slides down the plane, travels across a rough horizontal surface with coefficient of kinetic friction $μ$ , and compresses a spring with force constant $k$ a distance $x$ before momentarily coming to rest. Then the spring extends and the block travels back across the rough surface, sliding up the plane. The block travels a total distance $d$ on rough horizontal surface. The correct expression for the maximum height $h’$ that the block reaches on its return is

A block of mass $2\,\,kg$ is placed on a rough inclined plane as shown in the figure $(\mu = 0.2)$ so that it just touches the spring. The block is allowed to move downwards. The spring will be compressed to a maximum of

A chain of mass $m$ and length $l$ is hanging freely from edge $A$ (as shown in diagram $I$ ). Calculate the work done to fold it as shown in diagram $(II)$

A spring when stretched by $2 \,mm$ its potential energy becomes $4 \,J$. If it is stretched by $10 \,mm$, its potential energy is equal to