Question

A ideal spring of force constant k is connected to a small block of mass m using an inextensible light string (see fig). The pulley is mass less and friction coefficient between the block and the horizontal surface is $\mu = \frac{1}{\sqrt{3}}$. The string between the pulley and the block is vertical and has length l. Find the minimum velocity u that must be given to the block in horizontal direction shown, so that subsequently it leaves contact with the horizontal surface. [Take $k = \frac{2mg}{l}$]

Answer 1

$\sqrt{\frac{4gl}{\sqrt{3}} \ln(\sqrt{3}+2)}$

Answer 2

The problem asks for the minimum velocity 'u' that must be given to the block in the horizontal direction so that it subsequently leaves contact with the horizontal surface.

1. Condition for leaving contact:

The block leaves contact with the horizontal surface when the normal force (N) exerted by the surface on the block becomes zero.

2. Analyze the geometry and forces:

Initially, the string segment between the pulley and the block is vertical and has length 'l'. Let the block move horizontally by a distance 'x'.
The new length of the string segment from the pulley to the block will be $L' = \sqrt{l^2 + x^2}$.
Since the string is inextensible, the spring must have been stretched by an amount $\Delta L = L' - l = \sqrt{l^2 + x^2} - l$.
The tension in the string (T) is equal to the force exerted by the spring:
$T = k \Delta L = k (\sqrt{l^2 + x^2} - l)$.

Let $\theta$ be the angle the string (from pulley to block) makes with the vertical.
From the geometry, $\cos \theta = \frac{l}{\sqrt{l^2 + x^2}}$.

The forces acting on the block are:

• Gravitational force (mg) acting downwards.
• Normal force (N) acting upwards.
• Tension (T) acting along the string, at an angle $\theta$ to the vertical.
• Friction force ($f_k$) acting horizontally, opposing motion.

For vertical equilibrium (while the block is in contact):
$N + T \cos \theta - mg = 0$
So, $N = mg - T \cos \theta$.

When the block leaves contact, $N=0$.
Therefore, $mg - T \cos \theta = 0 \implies T \cos \theta = mg$.
Substitute $T = k (\sqrt{l^2 + x^2} - l)$ and $\cos \theta = \frac{l}{\sqrt{l^2 + x^2}}$:
$k (\sqrt{l^2 + x^2} - l) \frac{l}{\sqrt{l^2 + x^2}} = mg$
Given $k = \frac{2mg}{l}$:
$\frac{2mg}{l} (\sqrt{l^2 + x^2} - l) \frac{l}{\sqrt{l^2 + x^2}} = mg$
$2mg \left(1 - \frac{l}{\sqrt{l^2 + x^2}}\right) = mg$
Divide by mg:
$2 \left(1 - \frac{l}{\sqrt{l^2 + x^2}}\right) = 1$
$2 - \frac{2l}{\sqrt{l^2 + x^2}} = 1$
$\frac{2l}{\sqrt{l^2 + x^2}} = 1$
$\sqrt{l^2 + x^2} = 2l$
Square both sides:
$l^2 + x^2 = 4l^2$
$x^2 = 3l^2$
$x = \sqrt{3}l$

This is the horizontal displacement at which the block leaves contact.
At this point, $\cos \theta = \frac{l}{\sqrt{l^2 + (\sqrt{3}l)^2}} = \frac{l}{\sqrt{l^2 + 3l^2}} = \frac{l}{\sqrt{4l^2}} = \frac{l}{2l} = \frac{1}{2}$.
So, $\theta = 60^\circ$.
The tension at this point is $T = \frac{mg}{\cos \theta} = \frac{mg}{1/2} = 2mg$.
Also, the spring extension is $\Delta L = \sqrt{l^2+x^2}-l = 2l-l = l$.
Spring force $T = k \Delta L = \frac{2mg}{l} \cdot l = 2mg$. This is consistent.

3. Apply Work-Energy Theorem:

The initial kinetic energy of the block is $KE_i = \frac{1}{2}mu^2$.
The final state is when the block leaves contact at $x = \sqrt{3}l$.
The final potential energy stored in the spring is $PE_{s,f} = \frac{1}{2}k(\Delta L)^2 = \frac{1}{2}k(l)^2$.
Substitute $k = \frac{2mg}{l}$:
$PE_{s,f} = \frac{1}{2} \left(\frac{2mg}{l}\right) l^2 = mgl$.

The work done by friction ($W_f$) must be considered. The friction force is $f_k = \mu N$.
The normal force $N = mg - T \cos \theta$.
Using the expressions for T and $\cos \theta$:
$N = mg - k(\sqrt{l^2+x^2}-l) \frac{l}{\sqrt{l^2+x^2}}$
$N = mg - \frac{2mg}{l}(\sqrt{l^2+x^2}-l) \frac{l}{\sqrt{l^2+x^2}}$
$N = mg - 2mg \left(1 - \frac{l}{\sqrt{l^2+x^2}}\right)$
$N = mg - 2mg + \frac{2mgl}{\sqrt{l^2+x^2}}$
$N = -mg + \frac{2mgl}{\sqrt{l^2+x^2}}$

The work done by friction is $W_f = -\int_0^{\sqrt{3}l} f_k dx = -\int_0^{\sqrt{3}l} \mu N dx$.
$W_f = -\int_0^{\sqrt{3}l} \mu \left(-mg + \frac{2mgl}{\sqrt{l^2+x^2}}\right) dx$
$W_f = -\mu mg \int_0^{\sqrt{3}l} \left(-1 + \frac{2l}{\sqrt{l^2+x^2}}\right) dx$
The integral of $\frac{1}{\sqrt{a^2+x^2}}$ is $\ln|x+\sqrt{a^2+x^2}|$.
$W_f = -\mu mg \left[ -x + 2l \ln(x + \sqrt{l^2+x^2}) \right]_0^{\sqrt{3}l}$
Evaluate the limits:
At $x=\sqrt{3}l$: $-\sqrt{3}l + 2l \ln(\sqrt{3}l + \sqrt{l^2+(\sqrt{3}l)^2}) = -\sqrt{3}l + 2l \ln(\sqrt{3}l + 2l) = -\sqrt{3}l + 2l \ln(l(\sqrt{3}+2))$.
At $x=0$: $0 + 2l \ln(0 + \sqrt{l^2+0}) = 2l \ln(l)$.
So, $W_f = -\mu mg \left[ (-\sqrt{3}l + 2l \ln(l(\sqrt{3}+2))) - (2l \ln(l)) \right]$
$W_f = -\mu mg \left[ -\sqrt{3}l + 2l (\ln(l) + \ln(\sqrt{3}+2)) - 2l \ln(l) \right]$
$W_f = -\mu mg \left[ -\sqrt{3}l + 2l \ln(\sqrt{3}+2) \right]$
$W_f = \mu mg l (\sqrt{3} - 2 \ln(\sqrt{3}+2))$.

The minimum velocity 'u' means that the block just reaches the point of leaving contact ($x=\sqrt{3}l$) with zero kinetic energy. If it stops before this point, it won't leave contact. If it has kinetic energy at this point, 'u' would be larger than minimum.
So, at $x=\sqrt{3}l$, $KE_f = 0$.

Applying the Work-Energy Theorem: $KE_i + W_f = KE_f + PE_{s,f}$
$\frac{1}{2}mu^2 + \mu mg l (\sqrt{3} - 2 \ln(\sqrt{3}+2)) = 0 + mgl$
$\frac{1}{2}mu^2 = mgl - \mu mg l (\sqrt{3} - 2 \ln(\sqrt{3}+2))$
$\frac{1}{2}u^2 = gl - \mu gl (\sqrt{3} - 2 \ln(\sqrt{3}+2))$
$\frac{1}{2}u^2 = gl [1 - \mu (\sqrt{3} - 2 \ln(\sqrt{3}+2))]$
Substitute $\mu = \frac{1}{\sqrt{3}}$:
$\frac{1}{2}u^2 = gl \left[1 - \frac{1}{\sqrt{3}} (\sqrt{3} - 2 \ln(\sqrt{3}+2))\right]$
$\frac{1}{2}u^2 = gl \left[1 - 1 + \frac{2}{\sqrt{3}} \ln(\sqrt{3}+2)\right]$
$\frac{1}{2}u^2 = gl \left[\frac{2}{\sqrt{3}} \ln(\sqrt{3}+2)\right]$
$u^2 = \frac{4gl}{\sqrt{3}} \ln(\sqrt{3}+2)$
$u = \sqrt{\frac{4gl}{\sqrt{3}} \ln(\sqrt{3}+2)}$