Question

A spring is attached with a small block of mass $m$ and a fixed ceiling. The block is lying on a smooth horizontal table and initially the spring is vertical and unstretched. Natural length of the spring is $3l_0$. A constant horizontal force $F$ is applied on the block so that it moves in the direction of force. When length of the spring becomes $5l_0$, block is about to leave contact with the table, and its velocity at that instant is zero. Then ratio $\frac{mg}{F}$ is ______.

Answer 1

$\frac{12}{5}$

Answer 2

Solution Overview:

1. Geometry at the instant of leaving:

   • Ceiling-to-table distance = $3l_0$.
   • At the moment considered, spring length $= 5l_0$ so the extension is $5l_0-3l_0=2l_0$.
   • In the right‐triangle formed by the spring, vertical and horizontal arms, the vertical side $= 3l_0$ and hypotenuse $= 5l_0$. Thus, $$\cos\theta = \frac{3l_0}{5l_0} = \frac{3}{5} \quad\text{and}\quad \sin\theta=\frac{4l_0}{5l_0}=\frac{4}{5}.$$

2. Vertical Force Balance (Loss of Contact Condition):

   • At the instant the block is about to lose contact, the normal force $N=0$. The only vertical forces are from gravity ($mg$ downward) and the vertical component of the spring force (upward).
   • Spring force: $T=k\,\Delta l=k(2l_0)$.
   • Equate vertical forces for equilibrium: $$T\cos\theta = mg \quad \Longrightarrow \quad k(2l_0)\left(\frac{3}{5}\right)= mg.$$ Hence, $$k\,l_0 = \frac{5}{6}mg.$$

3. Work-Energy Consideration in Horizontal Direction:

   • Initially, the spring is unstretched and the block is at rest.
   • When the block moves, the horizontal displacement equals the horizontal side of the triangle, which is $4l_0$.
   • Since the block comes to rest at the final position, the work done by the force $F$ goes into stored spring energy: $$F\,(4l_0)=\frac{1}{2}k(2l_0)^2=\frac{1}{2}k(4l_0^2)=2k\,l_0^2.$$ Thus, $$F=\frac{2k\,l_0^2}{4l_0}=\frac{k\,l_0}{2}.$$
   • Substitute $k\,l_0=\frac{5}{6}mg$: $$F=\frac{1}{2}\left(\frac{5}{6}mg\right)=\frac{5}{12}mg.$$

4. Find $\frac{mg}{F}$:

   $$\frac{mg}{F}=\frac{mg}{\frac{5}{12}mg}=\frac{12}{5}.$$