Question

An object is attached to a vertical spring and slowly lowered to its equilibrium position. This stretches the spring by an amount d. If the same object is attached to the same vertical spring but permitted to fall instead. Through what distance does it stretch the spring?

• A. d
• B. 2d
• C. 3
• D. 5d

Answer 1

The correct option is B

2d

kd = mg

d = $\frac{mg}{k}$

So if i lower the block slowly the block stops after coming down by distance d

where d = $\frac{mg}{k}$ .. (i)

Now if I leave the block from natural lenght the block will reach distance d but it will have acquired a velocity so it will stop after falling a further distance , say $x_0$.

Let's apply work energy theorem initial velocity = 0 = final velocity.

$\Rightarrow$ K$E_i$ = K$E_f$= 0

Only forces acting on the block are spring force and gravity

$\Rightarrow$ $W_{sp}$ + $W_g$ = $\Delta$ kE = 0 ..(ii)

$mgx-\frac{1}{2}k{x}^2=0mgx=\frac{1}{2}k{x}^2\frac{2mg}{k}=x\Rightarrow x=2\frac{mg}{k}=2d$