A sphere of $2\,kg$ is released (assume $g=10\frac{m}{s^2}$) from a height of $2\,m$ with respect of a horizontal surface. If due air resistance $10\,J$ are dissipated by each meter that the sphere goes down. Find the mechanical energy with respect to the ground (in Joules) on the instant when the kinetic energy is equal to $0.5$ times its potential energy.

The alternatives given are as follows:

$\begin{array}{ll}

1.&350\,J\\

2.&300\,J\\

3.&250\,J\\

4.&200\,J\\

\end{array}$

I'm confused exactly how to tackle this problem.

Essentially what I believe is that:

$E_{k}=\frac{1}{2}E_{u}$

But the thing here that comes into play is that there's energy dissipated when the sphere is falling.

Then the mechanical energy would be as follows:

$E_{u}=E_{k}+E_{u'}$

At that given height there would be kinetic energy and potential energy.

But it mentions that the kinetic energy will be at that given height $\frac{1}{2}E_{u}$

$E_{u}-150=\frac{1}{2}E_{u}+E_{u'}$

$mg(2)-150=\frac{1}{2}(mg(2-x))+mg(2-x)$

By inserting the given values for the mass and gravity this is reduced to:

$x=\frac{17}{3}$

But that's where I became stuck. Can someone help me with this problem?