# Why equating forces give wrong answer here but using work-energy theorem gives the correcct answer? [closed]

Question: A block of Mass m is connected to another block of mass M by a mass-less spring of spring constant k . The blocks are kept on a smooth horizontal plane and are at rest. The spring is unstretched when a constant force F starts acting on the block of mass M(horizontally) to pull it. Find the maximum extension of the spring.

Here I assumed that max. extension(x) will be produced when both the blocks would be moving with constant and same acceleration. Then by considering the FBD diagram of the blocks m and M, I formed this equation

$$\Large \frac{kx}{m}= \frac{F-kx}{M}$$

and hence I get the value of x as

$$\Large x=\frac{mF}{k(m+M)}$$

$$\large Solution$$

by conservation of energy In the reference frame of center of mass.

For block 'm', we have two forces acting $$mF/(m+M)$$ and $$kx$$, in opposite directions.

For block 'M' we have three forces acting $$MF/(m+M)$$ and $$kx$$, and $$F$$ in the opposite direction.

Assuming m moves a max distance x_1 from CM M moves a distance x_2 from CM. Then work done by external force will be

$$W=mF(x_1+x_2)/(m+M)$$

This will be stored as the internal energy, therefore

$$1/2·k·(x_1+x_2)^2=W$$

On solving this we get

$$x_1+x_2=2mF/(k(m+M))$$

But my method is not correct. Can someone please help me out understand why this approach is not correct?

(When I looked up for the solution, the correct answer given was $$\large \frac{2mF}{k(m+M)}$$, by solving it from reference frame as COM, by using work-energy theorem but if that is the answer both the blocks would have different accelerations .)

• I think there is a typo in "Assuming m moves a max distance x_1 from CM M moves a distance x_2 from CM." – DKNguyen Nov 18 at 15:12