Question

Masters spring when stretched by a force 150 Newton show an extension of 1.5 metres the spring is arranged a block of mass 10 kg is released at rest on the frictionless inclined at an angle 30 degree it slide down and compressors the spring by 3 metres before coming to rest momentarily distance by which the blocks flight between or before hitting the spring​

Answer 1

Let $\sf{g=10\ m\,s^{-2}.}$

Since the spring shows an extension of $\sf{1.5\ m}$ on stretching it by a force $\sf{150\ N,}$ the spring constant is,

$\longrightarrow\sf{k=\dfrac{150}{1.5}}$

$\longrightarrow\sf{k=100\ N\,m^{-1}}$

Mass of the block, $\sf{m=10\ kg}$

Angle of inclination, $\sf{\theta=30^o}$

Normally the block slides down the inclined plane due to the acceleration,

$\longrightarrow\sf{a=g\sin\theta}$

$\longrightarrow\sf{a=10\sin30^o}$

$\longrightarrow\sf{a=5\ m\,s^{-2}}$

The block has this acceleration until it touches the spring.

Let the block have a velocity $\sf{v}$ when it touches the spring.

At the time when the block touches the spring, the block has kinetic energy, i.e., $\sf{\dfrac{1}{2}\,mv^2,}$ but the spring remains in normal state so it has no potential energy due to elasticity.

Hence the total energy of the system at this time is,

$\longrightarrow\sf{E_1=\dfrac{1}{2}\,mv^2}$

At the time when the spring gets compressed by $\sf{3\ m}$ and the block comes to rest, the block has no kinetic energy but the spring has maximum potential energy due to elasticity, i.e., $\sf{\dfrac{1}{2}\,kx^2.}$

Hence the total energy of the system at this time is,

$\longrightarrow\sf{E_2=\dfrac{1}{2}\,kx^2}$

By law of conservation of mechanical energy,

$\longrightarrow\sf{E_1=E_2}$

$\longrightarrow\sf{\dfrac{1}{2}\,mv^2=\dfrac{1}{2}\,kx^2}$

$\longrightarrow\sf{v^2=\dfrac{kx^2}{m}}$

$\longrightarrow\sf{v^2=\dfrac{100\times3^2}{10}}$

$\longrightarrow\sf{v^2=90}$

Let $\sf{s}$ be the distance travelled by the block from rest before touching the spring. Then by third equation of motion,

$\longrightarrow\sf{s=\dfrac{v^2-u^2}{2a}}$

Since the block is released from rest,

$\longrightarrow\sf{s=\dfrac{90-0^2}{2\times5}}$

$\longrightarrow\sf{\underline{\underline{s=9\ m}}}$

Hence the block travelled 9 m on the inclined plane before hitting the spring.