Question

Hii !

Solve it given in attachment .

Previous year JEE question . ​

Answer 1

On pulling the block by a force F, we know that a restoring force is developed in it, and is given by,

$F=-kx$

where x is the displacement of the block after being pulled.

We have to find the maximum velocity attained by the block when it is released after being pulled.

The acceleration of the block here,

$a=\dfrac{F}{m}\\\\\\a=-\dfrac {kx}{m}$

But,

$a=\dfrac {dv}{dt}\\\\\\a=\dfrac {dv}{dx}\cdot\dfrac {dx}{dt}\\\\\\a=\dfrac{dv}{dx}\cdot v$

Then,

$\dfrac {dv}{dx}\cdot v=-\dfrac {kx}{m}\\\\\\v\ dv=-\dfrac {k^2}{mk}\cdot x\ dx$

Now it's time for integration.

→ The velocity (LHS) is integrated from 0 (since the block is released from rest) to the maximum velocity $v$.

→ The maximum velocity is attained when the block moves the same distance x backwards, i.e., in the opposite direction of the pulling. So the displacement (RHS) is integrated from x (since the block is released after being pulled by x) to 0.

So,

$\displaystyle\int\limits_0^vv\ dv=-\dfrac {k^2}{mk}\int\limits_x^0x\ dx\\\\\\\left [\dfrac {v^2}{2}\right]_0^v=-\dfrac {k^2}{mk}\left [\dfrac {x^2}{2}\right]_x^0\\\\\\\dfrac {v^2}{2}=\dfrac {k^2x^2}{2mk}\\\\\\v^2=\dfrac {F^2}{mk}\\\\\\\underline {\underline {v=\dfrac {F}{\sqrt{mk}}}}$

Hence (a) is the answer.

Answer 2

Answer:

I THINK ITS (A) F/√MK

Explanation:

sorry