Difficulty in deciding when to apply work-energy theorem [closed]

Question

I was trying to do a certain question in physics and the answer I was obtaining was different than the actual answer I don't know why?
Two blocks A and B of the same mass connected with a spring are placed on a rough inclined plane, which makes an angle $\theta$ with horizontal. What minimum velocity should be given to A up the incline so that B just moves?
This is how I tried to do it. The force required to move B up the incline is $kx$ where x is elongation and k is spring constant. we know that spring force is greater than $mg(sin\theta+\mu cos\theta)$. And we can use work-energy theorem to figure out velocity. $0.5*k*x^2=0.5*mv^2$ where $0.5*k*x^2$ is work done by spring force. and when you count all the chickens $v$ turns out to be $\sqrt{km}(gsin\theta+\mu gcos\theta)$. Which apparently is the wrong answer. And the correct answer apparently is $$\sqrt{(3m)/k}(gsin\theta+\mu gcos\theta).$$ I have no idea what I did wrong. Can somebody help? is there something wrong with the WOrk energy theorem, or what?

Answer 1

If the body is just about to move, then the net force on body(b) is zero, we just need a little push. Also all the initial kinetic energy of body(a) will be used, since body(b) in the end is very close to rest. $$mgsin{\theta}+\mu{mg}cos{\theta}=kx$$ Using work-energy principle. $$\frac{1}{2}mv_{i}^2=\sum{W}$$ $$\frac{1}{2}mv_{i}^2=mgsin{\theta}x+\frac{1}{2}kx^2+\mu{mg}cos{\theta}x$$ $$\frac{1}{2}mv_{i}^2=\frac{1}{2}kx^2+kx^2$$ $$v_{i}=x\sqrt{\frac{3k}{m}}$$ Substitute x and rest is solvable.