Question

A block of mass $M$ is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force constant value $k$. The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be

• A. 2Mg/k
• B. 4Mg/k
• C. Mg/2k
• D. Mg/k

Answer 1

The Correct Option is A

To determine the maximum extension produced in the length of the spring, we will analyze the energy transformations involved in the system.

Initially, the block is released from rest when the spring is unstretched, meaning that the potential energy stored in the spring is zero, and the block has only gravitational potential energy.

Let's denote the following:

• Mass of the block $M$
• Gravitational acceleration $g$
• Spring constant $k$
• Maximum extension of the spring $x_{\text{max}}$

At the maximum extension, the block momentarily comes to rest. At this point, all the gravitational potential energy has been converted to elastic potential energy stored in the spring.

The initial gravitational potential energy when the block falls by a distance $x_{\text{max}}$ is given by:

U_g = Mgx_{\text{max}}

The elastic potential energy stored in the spring at maximum extension is:

U_s = \frac{1}{2} k x_{\text{max}}^2

By conservation of energy, initial gravitational potential energy equals the elastic potential energy at maximum extension:

Mgx_{\text{max}} = \frac{1}{2} k x_{\text{max}}^2

To find $x_{\text{max}}$, we rearrange the equation:

kx_{\text{max}}^2 = 2Mgx_{\text{max}}

Divide both sides by $x_{\text{max}}$ (assuming $x_{\text{max}}\neq 0$):

kx_{\text{max}} = 2Mg

Solve for $x_{\text{max}}$:

x_{\text{max}} = \frac{2Mg}{k}

Therefore, the maximum extension produced in the spring is \frac{2Mg}{k}.

Thus, the correct answer is 2Mg/k.