Question

$\Lambda$ small bar $\Lambda$ resting on a smooth, horizontal surface is attached by threads to the wall and by means of a massless pullcy, to a block $B$ of the same mass as the bar itself. Further, the bar is attached to a point $O$ by means of a light unstretched spring of lenght $l_{0}$ and spring constant $k=5 m g /_{0}$, where $m$ is mass of the bar. When the thread is burnt, the bar starts moving. Find the velocity when $\Lambda$ breaks off the surface. Solution Let us consider the instantaneous diplacement of $A$ to be $x$ and the spring be inclined to the vertical by $\theta$. spring force $F-k \sqrt{l_{*}^{2} \mid x^{2}} \quad l_{o}$ Trom figurc (a), $T-k\left(\sqrt{l_{0}^{2}+x^{2}}-l_{a}\right) \sin \theta-m \frac{d v}{d t}$ and $N+k\left(\sqrt{l_{0}^{2} \mid x^{2}} \quad l_{0}\right) \cos \theta-m g$ From figure (b), $m g \quad T-m \frac{d v}{d t}$ Combining (i) and (iii) $m g \quad k\left(\sqrt{l_{0}^{2} \mid x^{2}}-l_{0}\right) \sin \theta-2 m \frac{d v}{d t}$ When $A$ breaks of the planc, $N=0$ $\therefore k\left(\sqrt{l_{n}^{2}+x^{2}}-l_{0}\right) \cos \theta=m g$ or $\frac{5 m g}{l_{n}}\left(\sqrt{l_{n}^{2}+x^{2}}-l\right) \frac{l_{n}}{\sqrt{l_{0}^{2} \mid x^{2}}}-m g$ or $x-\frac{3}{4} l \quad\left(\because k-\frac{5 m g}{l_{o}}\right.$ (given) and $\cos \theta-\frac{l_{o}}{\sqrt{l_{n}^{2} \mid x^{2}}}$ from the figure) \mathrm{\{} T r o m ~ t h e ~ p r i n c i p l e ~ o f ~ c o n s e r v a t i o n ~ o f ~ c n e r g y ~ $-m g h-2\left(\frac{1}{2} m v^{2}\right)+\frac{1}{2} k \Delta l^{2}-m g(h+x)$ where $h$ is the initial distance of $B$ below the table and $\Delta l$ is the increase in length. When $A$ breaks of $x-\frac{3}{4} l$$\therefore \quad A-\sqrt{l_{0}^{2}+\frac{9}{16} l_{0}}-l_{o}-\frac{l_{o}}{4} \quad \therefore \quad-m g h-m v^{2}+\frac{1}{2} k \frac{l_{o}^{2}}{16}-m g\left(h+\frac{3 l_{v}}{4}\right)$ or $m g \frac{3 l_{o}}{4}-m v^{2}+\frac{1}{2} \frac{5 m g}{l_{o}} \frac{l_{o}^{2}}{16} \quad$ or $\quad \frac{3 g l_{o}}{4}-m v^{2}+\frac{5 g l_{o}}{16 \times 2}$ or $v^{2}-\frac{3}{4} g l, \frac{5 g l_{o}}{32} \times \frac{19 g l_{n}}{32}$ $\therefore \quad v-\sqrt{\frac{19 g l_{n}}{32}}$

   $\Lambda$ small bar $\Lambda$ resting on a smooth, horizontal surface is attached by threads to the wall and by means of a massless pullcy, to a block $B$ of the same mass as the bar itself. Further, the bar is attached to a point $O$ by means of a light unstretched spring of lenght $l_{0}$ and spring constant $k=5 m g /_{0}$, where $m$ is mass of the bar. When the thread is burnt, the bar starts moving. Find the velocity when $\Lambda$ breaks off the surface.
Solution Let us consider the instantaneous diplacement of $A$ to be $x$ and the spring be inclined to the vertical by $\theta$. spring force $F-k \sqrt{l_{*}^{2} \mid x^{2}} \quad l_{o}$
Trom figurc (a), $T-k\left(\sqrt{l_{0}^{2}+x^{2}}-l_{a}\right) \sin \theta-m \frac{d v}{d t}$
and $N+k\left(\sqrt{l_{0}^{2} \mid x^{2}} \quad l_{0}\right) \cos \theta-m g$
From figure (b), $m g \quad T-m \frac{d v}{d t}$
Combining (i) and (iii) $m g \quad k\left(\sqrt{l_{0}^{2} \mid x^{2}}-l_{0}\right) \sin \theta-2 m \frac{d v}{d t}$
When $A$ breaks of the planc, $N=0$
$\therefore k\left(\sqrt{l_{n}^{2}+x^{2}}-l_{0}\right) \cos \theta=m g$
or $\frac{5 m g}{l_{n}}\left(\sqrt{l_{n}^{2}+x^{2}}-l\right) \frac{l_{n}}{\sqrt{l_{0}^{2} \mid x^{2}}}-m g$
or $x-\frac{3}{4} l \quad\left(\because k-\frac{5 m g}{l_{o}}\right.$ (given) and $\cos \theta-\frac{l_{o}}{\sqrt{l_{n}^{2} \mid x^{2}}}$ from the figure)
\mathrm{\{} T r o m ~ t h e ~ p r i n c i p l e ~ o f ~ c o n s e r v a t i o n ~ o f ~ c n e r g y ~
$-m g h-2\left(\frac{1}{2} m v^{2}\right)+\frac{1}{2} k \Delta l^{2}-m g(h+x)$
where $h$ is the initial distance of $B$ below the table and $\Delta l$ is the increase in length.
When $A$ breaks of $x-\frac{3}{4} l$$\therefore \quad A-\sqrt{l_{0}^{2}+\frac{9}{16} l_{0}}-l_{o}-\frac{l_{o}}{4} \quad \therefore \quad-m g h-m v^{2}+\frac{1}{2} k \frac{l_{o}^{2}}{16}-m g\left(h+\frac{3 l_{v}}{4}\right)$
or $m g \frac{3 l_{o}}{4}-m v^{2}+\frac{1}{2} \frac{5 m g}{l_{o}} \frac{l_{o}^{2}}{16} \quad$ or $\quad \frac{3 g l_{o}}{4}-m v^{2}+\frac{5 g l_{o}}{16 \times 2}$
or $v^{2}-\frac{3}{4} g l, \frac{5 g l_{o}}{32} \times \frac{19 g l_{n}}{32}$
$\therefore \quad v-\sqrt{\frac{19 g l_{n}}{32}}$

Practice Problem in Physics for the JEE Main and Advanced

Abhay Kumar 2015 Edition

Chapter 5, Problem 32

Instant Answer

Solved by Expert Vivek Singh

01/31/2022

Step 1

The spring force $F$ can be written as $F=k \sqrt{l_{0}^{2} + x^{2}} - l_{0}$. Show more…

Show all steps

Thanks for your feedback!

$\Lambda$ small bar $\Lambda$ resting on a smooth, horizontal surface is attached by threads to the wall and by means of a massless pullcy, to a block $B$ of the same mass as the bar itself. Further, the bar is attached to a point $O$ by means of a light unstretched spring of lenght $l_{0}$ and spring constant $k=5 m g /_{0}$, where $m$ is mass of the bar. When the thread is burnt, the bar starts moving. Find the velocity when $\Lambda$ breaks off the surface. Solution Let us consider the instantaneous diplacement of $A$ to be $x$ and the spring be inclined to the vertical by $\theta$. spring force $F-k \sqrt{l_{*}^{2} \mid x^{2}} \quad l_{o}$ Trom figurc (a), $T-k\left(\sqrt{l_{0}^{2}+x^{2}}-l_{a}\right) \sin \theta-m \frac{d v}{d t}$ and $N+k\left(\sqrt{l_{0}^{2} \mid x^{2}} \quad l_{0}\right) \cos \theta-m g$ From figure (b), $m g \quad T-m \frac{d v}{d t}$ Combining (i) and (iii) $m g \quad k\left(\sqrt{l_{0}^{2} \mid x^{2}}-l_{0}\right) \sin \theta-2 m \frac{d v}{d t}$ When $A$ breaks of the planc, $N=0$ $\therefore k\left(\sqrt{l_{n}^{2}+x^{2}}-l_{0}\right) \cos \theta=m g$ or $\frac{5 m g}{l_{n}}\left(\sqrt{l_{n}^{2}+x^{2}}-l\right) \frac{l_{n}}{\sqrt{l_{0}^{2} \mid x^{2}}}-m g$ or $x-\frac{3}{4} l \quad\left(\because k-\frac{5 m g}{l_{o}}\right.$ (given) and $\cos \theta-\frac{l_{o}}{\sqrt{l_{n}^{2} \mid x^{2}}}$ from the figure) \mathrm{\{} T r o m ~ t h e ~ p r i n c i p l e ~ o f ~ c o n s e r v a t i o n ~ o f ~ c n e r g y ~ $-m g h-2\left(\frac{1}{2} m v^{2}\right)+\frac{1}{2} k \Delta l^{2}-m g(h+x)$ where $h$ is the initial distance of $B$ below the table and $\Delta l$ is the increase in length. When $A$ breaks of $x-\frac{3}{4} l$$\therefore \quad A-\sqrt{l_{0}^{2}+\frac{9}{16} l_{0}}-l_{o}-\frac{l_{o}}{4} \quad \therefore \quad-m g h-m v^{2}+\frac{1}{2} k \frac{l_{o}^{2}}{16}-m g\left(h+\frac{3 l_{v}}{4}\right)$ or $m g \frac{3 l_{o}}{4}-m v^{2}+\frac{1}{2} \frac{5 m g}{l_{o}} \frac{l_{o}^{2}}{16} \quad$ or $\quad \frac{3 g l_{o}}{4}-m v^{2}+\frac{5 g l_{o}}{16 \times 2}$ or $v^{2}-\frac{3}{4} g l, \frac{5 g l_{o}}{32} \times \frac{19 g l_{n}}{32}$ $\therefore \quad v-\sqrt{\frac{19 g l_{n}}{32}}$

Play audio

Feedback

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

System Geometry in Spring Extensions

Geometric relationships are fundamental when analyzing systems involving springs and inclined forces. In such cases, the extension of a spring or the components of forces can often be found using trigonometric and Pythagorean relationships, linking linear displacement to angular orientations. This allows a proper formulation of the forces involved in the motion of bodies connected by springs or threads.

Hooke's Law and Elastic Potential Energy

Hooke's law describes how the force exerted by a spring is proportional to its extension or compression from its equilibrium length. The potential energy stored in a spring is determined by this elastic force and is a key consideration when analyzing systems where spring forces are significant, particularly when converting between different forms of mechanical energy.

Force Equilibrium and Normal Force

When analyzing bodies in contact with surfaces, the normal force is the reactive force exerted by the surface. Assessing conditions such as when the normal force drops to zero is important, as it indicates a change in the regime of motion (for example, when a body loses contact with a supporting surface), and is critical for setting up the correct boundary conditions in dynamic problems.

Conservation of Energy

This principle states that in an isolated system, the total energy remains constant, though it may change forms, such as from potential to kinetic or elastic potential energy. In mechanics, it is frequently used to relate the work done by forces (including gravitational and spring forces) to changes in an object's speed or configuration.

Newton's Second Law of Motion

Newton's second law establishes that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. This law is essential for setting up equations that link the applied forces, the resulting acceleration, and consequently the motion of the body under study.

Please give Ace some feedback