Question

A block of mass $m$, lying on a smooth horizontal surface, is attached to a spring (of negligible mass) of spring constant $k$. The other end of the spring is fixed, as shown in the figure. The block is initally at rest in its equilibrium position. If now the block is pulled with a constant force $F$, the maximum speed of the block is :

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

Answer 1

The Correct Option is C

To find the maximum speed of the block, we first need to understand the system's behavior. Initially, the block is at rest and is pulled by a constant force \( F \). As the block is displaced, the spring compresses or extends, storing potential energy, and a point is reached where the block has maximum speed.

When the block reaches maximum speed, all the work done by the force \( F \) is converted into kinetic energy of the block and potential energy of the spring. Let's go through the steps to solve this:

1. The work done by the force \( F \) as the block moves to a point of maximum speed is given by: \[ W = F \times x \] where \( x \) is the displacement of the block at maximum speed.
2. At maximum speed, the kinetic energy (\( KE \)) of the block is: \[ KE = \frac{1}{2} m v^2 \] where \( v \) is the maximum speed.
3. The potential energy (\( PE \)) stored in the spring at that point is: \[ PE = \frac{1}{2} k x^2 \]
4. According to the work-energy principle, the work done by the force is equal to the total mechanical energy at maximum speed: \[ F \times x = \frac{1}{2} m v^2 + \frac{1}{2} k x^2 \]
5. To find the condition for maximum speed, take the derivative of the total energy with respect to \( x \) and set it to zero. Solving gives the relationship: \[ F = kx \] Substituting this into the energy equation: \[ F^2 = k^2 x^2 \quad \Rightarrow \quad x^2 = \left(\frac{F}{k}\right)^2 \]
6. Thus, substituting \( x = \frac{F}{k} \) \[ F \times \frac{F}{k} = \frac{1}{2} m v^2 + \frac{1}{2} k \left(\frac{F}{k}\right)^2 \] Simplifying: \[ \frac{F^2}{k} = \frac{1}{2} m v^2 + \frac{F^2}{2k} \]
7. Rearranging terms, solve for \( v \): \[ \frac{F^2}{2k} = \frac{1}{2} m v^2 \] \[ F^2 = mk v^2 \] \[ v^2 = \frac{F^2}{mk} \] \[ v = \frac{F}{\sqrt{mk}} \]

Therefore, the maximum speed of the block is \(\frac{F}{\sqrt{mk}}\). This matches with the given correct option:

\(\frac{F}{\sqrt{mk}}\)