Question

A massless spring when stretched by a force 150N
shows an extension of 1.5m. This spring is
arranged as shown in figure. A block of mass 10kg
is released at rest on a frictionless incline. It slides
down and compresses the spring by 3m before
coming to rest momentarily. Distance by which the
block slides before hitting the spring is
(1) 4m
(3) 9m
(2) 6m
(4) 12m​

Answer 1

Answer:

Thats the answer I guess..But ,u didn't give the Angle of inclination...

Answer 2

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.