Question

A block of mass $m$ is connected to a spring of force constant $k$. Initially the block is at rest and the spring has natural length. A constant force $F$ is applied horizontally towards right. The maximum speed of the block will be (there is no friction between block and the surface)

• A. $\frac{F}{\sqrt{2mk}}$
• B. $\frac{F}{\sqrt{mk}}$
• C. $\frac{\sqrt{2}F}{\sqrt{mk}}$
• D. $\frac{2F}{\sqrt{mk}}$

Answer 1

Answer: (B)

$\frac{F}{\sqrt{mk}}$

Maximum speed is at equilibrium where
$F=kx\Rightarrow x=\frac{F}{k}$

Now, $F\cdot x=\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2$

or $F\frac{F}{k}=\frac{1}{2}m{v}^2+\frac{1}{2}k{\frac{F}{k^2}}^2$

Solving we get,
$v=\frac{F}{\sqrt{mk}}=v_{max}$