Question

A constant force produces maximum velocity $v$ on the block connected to the spring of force constant $k$ as shown in the figure. When the force constant of spring becomes $4k$, the maximum velocity of the block is (block is at rest when spring is relaxed):

• A. $v/4$
• B. $2v$
• C. $v/2$
• D. $v$

Answer 1

Answer: (C)

By work energy theorem;
$F{x}_1-\frac{1}{2}k{x}_1^2=\frac{1}{2}m{v}^2...(1)$
and $F{x}_2-\frac{1}{2}{k}^{\prime }{x}_2^2=\frac{1}{2}m{v}^{\prime 2}....(2)$
where $x_1,x_2$ are initial and final extensions and $v,v^{\prime }$ are initial and final velocities.
In both cases, force applied is same, and velocity becomes maximum when $F=kx$. (after which the mass will decelerate)
$\therefore F=k{x}_1=(4k)x_2$
$\Rightarrow x_2=\frac{x_1}{4}$
Substituting in $(2)$:
$F\left(\frac{x_1}{4}\right)-\frac{1}{2}(4K){\left(\frac{x_1}{4}\right)}^2=\frac{1}{2}m{v}^2$
$\Rightarrow \frac{1}{4}\left[F{x}_1-\frac{1}{2}k{x}_1^2\right]=\frac{1}{2}m{v}^{\prime 2}...(3)$
Dividing $(3)/(1)$, we get:
$\frac{1}{4}=\frac{v^{\prime 2}}{v^2}$
$\Rightarrow v^{\prime }=\frac{v}{2}$