Question

Initially, the block of mass $M$ is at rest on a frictionless floor and the spring is in relaxed condition A constant force is applied on the block as shown in the figure. The maximum velocity of the block is:

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

Answer 1

The correct option is A

$\frac{F}{\sqrt{mK}}$

Solution:

Step 1: Given information

Mass $=M$

Surface $=$Frictionless

Step 2: Formula for maximum velocity

Maximum speed is at equilibrium where,

$$\begin{gathered} \Rightarrow F=kx \\ \Rightarrow x=\frac{F}{k} \end{gathered}$$

Where,

$F=$Force

$x=$Relative force

$k=$Force constant

Step 3: Calculate the maximum velocity of the block

​Now,$F\cdot x=\left(\frac{1}{2}\right)m{v}^2+\left(\frac{1}{2}\right)k{x}^2$

$F\cdot \left(\frac{F}{k}\right)=\left(\frac{1}{2}\right)m{v}^2+\left(\frac{1}{2}\right)k{\left(\frac{F}{k}\right)}^2$

Solving we get,

$v=\frac{F}{\sqrt{mK}}$.

Hence, the correct option is (A).