As others have said, your load resistor should go to ground, not to the inverting input. Looks like you have fixed the power supply issues as of this post.
The resistor ratio is correct, however your resistor values are WAY too low. The 200 ohm Rf (in your current schematic) goes to a virtual ground, so it's effectively in parallel with the load resistance of 2K as far as the op-amp output is concerned. It should give you some output, but the current limiting will kick in way before you get to 10V on the output. 10V out into 2K\$\Omega\$ || 200\$\Omega\$ = 55mA. That's why you are seeing "clipping" (or at least it should be, depending on how good your model is).
Use something more like 4K/20K and you should get a much better result. If you want to balance offset due to input bias current offset, use a resistor from the non-inverting input to ground.
With sensible resistors and 2V RMS in (2V peak => \$\sqrt 2 V_{RMS} = 1.414 V_{RMS}\$ = would be better since it looks like it's just starting to clip):
With given resistors and 2V RMS in (the cut-off number on the left is -10V, so the vertical scale is different from the other graph).
The reverse curve when 'clipping' is because the input is actually driving the output into the opposite direction (non-inverting) against the output (which is limiting and thus in a kind of constant-current mode).
You can usually find a fabrication hint on the front page of a device's data sheet - it might say something like: -
Single Supply, Rail-to-Rail, Low Power, FET-Input Op Amp
This is for the AD824 and it does mentions "FETs". However, I'd be more interested in it being low-power and rail-to-rail rather than the technology used.
But, "FET" input usually means low-bias currents so, do I then choose it on that basis? No, I certainly do not - I read the data sheet and look at the specified bias and offset currents and make a judgement on those numbers. In fact I don't give any credence to the word "FET" at all - I read the numbers and make a judgement on those numbers.
I did a search of "FET" in the document and a lot of mentions came up on page 1 (the BS page that should only be read with a pinch of salt) then, it wasn't until page 11 that the next mention of FET arose. One mention on page 12 and this time it said JFET - a more meaningful statement about technology but still, I'm not swayed because I read the numbers in the tables.
So, my advice from nearly 40 years of using op-amps on a regular basis is, forget the tech and read the data - understand the implications of each line in the tables of numbers because this is the true performance guarantee.
The other think is that if you go to TI's (or ADI's or LT's) website and do a search, it's a parametric search and not a fab/tech search. This should tell anyone to forget about fabrication methods and concentrate on real maximums and minimums and get reading the nitty-gritty of the data sheet.
However, if you don't know what you are aiming for as a design, spend more time thinking about this THEN start trawling the parametric tables and THEN compare data sheets.
Best Answer
I am no expert at motor control, but I think you are on the wrong track in a few ways:
An op-amp is probably not the best way to do this, but if you want to do it that way, using a 120 MHz op-amp for a circuit that probably needs a few 10's of Hz of bandwidth is probably not the right choice. You could probably do this with a $0.05 LM2904B (with a transistor buffer on the output).
Current feedback op-amps were designed to obtain higher frequency performance from older manufacturing processes. They have some design quirks and require special design consideration. But since you don't need 100 MHz or higher bandwidth, you can avoid the whole problem by using an ordinary voltage-feedback op-amp.
Since your output voltage is always above 0 V, you don't need a negative supply.
But in any case, the op-amp design will be fairly inefficient, burning power as heat to reduce the supply voltage to whatever voltage you are delivering to the motor. A more efficient solution, if you don't need exceptional accuracy in the voltage divide ratio from the input to the motor, would just use a PWM circuit, with 50% duty cycle. If the motor can't handle a PWM drive directly (this is where you need someone who knows motors better than me), then you can filter the signal with an LC filter.
simulate this circuit – Schematic created using CircuitLab