Conservation of energy on nail problems

Question

Suppose a spring is on a surface. An object of mass $m$ is in $h$ meter height from the top of the spring. When the object falls freely, the spring contracts by $x$ meter. Find $x$ when $k$ is spring constant.

The solution given in books is $mg(h+x)=\frac{1}{2}kx^2$.

Now,let us consider the top of the spring as reference plane. So potential energy at that plane is $0$. Now energy of the system at the beginning is only the potential energy of the object. So $E_1=mgh$

After the fall,the spring contracts by $x$ meter, so the potential energy of the spring is now $\frac{1}{2}kx^2$ and the potential energy of the object is $-mgx$ and kinetic energy of the object is

$\frac{1}{2}mv^2$.

So by conservation of energy,we get

$mgh=-mgx+\frac{1}{2}mv^2+\frac{1}{2}kx^2$.

Where did I go wrong?

Answer 1

Your only error is including the term $\frac{1}{2}mv^2$. The kinetic energy of the object has gone into the potential energy of the spring ($\frac{1}{2}mv^2$), i.e. $v$ is 0 once the spring is maximally compressed.