Question

The block will leave the horizontal surface if it can reach a point B where

$\qquad T \cos\theta = mg \ldots\ldots\ldots\ldots(i) \begin{bmatrix}T = \text{tension in string} \\ = \text{tension in string} \end{bmatrix}$

But $T = kx = kl \left[ \sec\theta - 1 \right] \left[ x = \text{extension in spring} = l(\sec\theta - 1) \right]$

Using (i) $\therefore kl (\sec\theta - 1)\cos\theta = mg \quad \left[ \text{at } B \right]$

4.46 PROBLEMS IN PHYSICS FOR JEE ADVANCED

$\Rightarrow 2mg(1-\cos\theta) = mg \Rightarrow 1 - \cos\theta = \frac{1}{2}$

$\Rightarrow \theta = 60^\circ$

From work energy theorem

Work done by friction = $\Delta k + \Delta U$

$-\mu mgl \tan 60^\circ = -\frac{1}{2}mu^2 + \frac{1}{2}k \left[ l(\sec 60^\circ - 1) \right]^2$

$\therefore \frac{1}{2}mu^2 = \frac{1}{2}kl^2 + \sqrt{3} \mu mgl. = \frac{1}{2}\frac{2mg}{l}.l^2 + \sqrt{3} \frac{1}{\sqrt{3}} mgl = 2mgl.$

$\therefore u = 2 \sqrt{gl}$

Answer 1

u = 2\sqrt{gl}

Answer 2

The problem describes a block on a horizontal surface connected to a spring via a pulley. The block moves from an initial position A to a point B, where it just leaves the horizontal surface. We need to determine the initial velocity 'u' required for this to happen, based on the provided solution steps.

Explanation of the Solution:

1. Condition for Leaving the Horizontal Surface: The block leaves the horizontal surface when the normal force acting on it becomes zero. At point B, the forces acting vertically on the block are its weight ($mg$) downwards and the vertical component of the tension ($T \cos\theta$) upwards. For the block to just lift off, these forces must balance: $T \cos\theta = mg \quad \ldots(i)$

2. Tension in the String and Spring Extension: The tension $T$ in the string is due to the extension of the spring. From the geometry, if the initial vertical height of the pulley above point A is $l$, then when the block is at point B, the length of the string from the pulley to B is $l/\cos\theta = l \sec\theta$. The extension $x$ in the spring is the difference between this new length and the initial length $l$: $x = l \sec\theta - l = l(\sec\theta - 1)$ According to Hooke's Law, the tension in the string (and thus the spring force) is $T = kx$, where $k$ is the spring constant: $T = kl(\sec\theta - 1)$

3. Determining the Angle $\theta$ at Lift-off: Substitute the expression for $T$ into equation (i): $kl(\sec\theta - 1)\cos\theta = mg$ $kl(1 - \cos\theta) = mg$ The solution then uses the relation $kl = 2mg$. This implies that the spring constant $k$ is $2mg/l$, which must be a given parameter for this specific problem. Substituting $kl = 2mg$: $2mg(1 - \cos\theta) = mg$ $1 - \cos\theta = \frac{1}{2}$ $\cos\theta = \frac{1}{2}$ This gives the angle $\theta = 60^\circ$.

4. Applying the Work-Energy Theorem: The work-energy theorem states that the work done by non-conservative forces ($W_{nc}$) equals the change in kinetic energy ($\Delta K$) plus the change in potential energy ($\Delta U$). Here, the non-conservative force is friction. $W_{friction} = \Delta K + \Delta U_{spring}$

   • Change in Kinetic Energy ($\Delta K$): The block starts with initial velocity $u$ (kinetic energy $K_i = \frac{1}{2}mu^2$) and just reaches point B, implying its final kinetic energy at B is zero ($K_f = 0$). $\Delta K = K_f - K_i = 0 - \frac{1}{2}mu^2 = -\frac{1}{2}mu^2$
   • Change in Spring Potential Energy ($\Delta U_{spring}$): The spring is initially unextended (or its extension is zero) at point A, so $U_{spring,i} = 0$. At point B, the extension is $x = l(\sec\theta - 1) = l(\sec 60^\circ - 1) = l(2 - 1) = l$. $U_{spring,f} = \frac{1}{2}kx^2 = \frac{1}{2}k l^2$ $\Delta U_{spring} = U_{spring,f} - U_{spring,i} = \frac{1}{2}kl^2 - 0 = \frac{1}{2}kl^2$
   • Work Done by Friction ($W_{friction}$): The horizontal distance moved by the block from A to B is $l \tan\theta$. At $\theta = 60^\circ$, this distance is $l \tan 60^\circ = \sqrt{3}l$. The problem statement provides the work done by friction as $-\mu mgl \tan 60^\circ$. This implies that for the friction calculation, the normal force is effectively taken as $mg$, or this is a given work value. $W_{friction} = -\mu mgl \tan 60^\circ = -\mu mgl \sqrt{3}$

5. Solving for Initial Velocity 'u': Substitute these values into the work-energy equation: $-\mu mgl \sqrt{3} = -\frac{1}{2}mu^2 + \frac{1}{2}kl^2$ Rearrange to find $\frac{1}{2}mu^2$: $\frac{1}{2}mu^2 = \frac{1}{2}kl^2 + \mu mgl \sqrt{3}$ Now, substitute the values $k = \frac{2mg}{l}$ and $\mu = \frac{1}{\sqrt{3}}$ (another given parameter for the problem): $\frac{1}{2}mu^2 = \frac{1}{2}\left(\frac{2mg}{l}\right)l^2 + \left(\frac{1}{\sqrt{3}}\right)mgl\sqrt{3}$ $\frac{1}{2}mu^2 = mgl + mgl$ $\frac{1}{2}mu^2 = 2mgl$ $mu^2 = 4mgl$ $u^2 = 4gl$ $u = 2\sqrt{gl}$