In the given figure, two blocks A and B, each having a mass of 320 grams are connected by a light string passing over a smooth light pulley. The horizontal surface on which block A can slide is smooth. The block A is attached to a spring of spring constant 40$\frac{N}m$ whose other end is fixed to a support 40cm above the horizontal surface. Initially, the spring is vertical and unstretched when the system is released to move. Find the velocity of the block A at the instant it breaks off the surface below it. Take g=10$\frac{m}{s^2}$.
What I have got till now:
After drawing the free body diagram of the block at the instant it breaks off the surface, I found these pieces of data:
The block breaks off when the smaller angle of spring with the ground is $\sin^{-1}\frac{4}5$
The distance that the block covers till this point is 0.3m
The elongation in the spring is equal to 0.1m
The part that I am confused with:
Now, according to what I have learned, the work done by any force on an object is $\vec{F}•d\vec{r}$. So this means that we have to take the varying angle between the spring and the displacement of the block into account while calculating the work done by it. So, the work done by spring should be:
$\int\ k\Delta s\cos\theta dx$
Here, $\Delta s$ is the elongation of the spring, $\theta$ is the smaller angle made by the spring with the ground, and we are integrating in terms of the displacement of the block.
After a bit of observation, I found that $dx$ is equal to -0.4$d\theta$. and $\Delta s$ is equal to 0.4$\frac{1-sin\theta}{sin\theta}$. Now, we can integrate in terms of $\theta$ and find the work done by the spring, which comes to about 2.32
Then, we can factor in the potential energy and find the kinetic energy from that.
But this is not the solution that I found when I searched through the internet. A couple of solutions I found online:
https://www.toppr.com/ask/question/figure-shows-two-blocks-a-and-b-each-having-a-mass-of-320-gram-connected/
https://www.sarthaks.com/44038/figure-8-e12-shows-two-blocks-a-and-b-each-having-a-mass-of-320-g-connected-by-a-light
At the end of both the solutions, the work done by spring is given as $\frac{1}2kx^2$. Which should not be correct as the force is not acting parallel or anti parallel to displacement.
The final answer is correct:$1.5\frac{m}{s}$, and the one I got is wrong. But, it does not make sense to me to put the work done by spring as $\frac{1}2kx^2$.
So, am I making a very stupid mistake here, or is there something else going on?
Question from: HC Verma- Concepts of Physics Part 1 Chapter 8 Q.50