I have an electric 180w motor from an electric go cart which currently runs on lead acid batteries but I want to convert it to have a 18650 battery pack for a power supply but I cant find the rated voltage and amperage for the motor I only know that it runs at 180w so can I use whatever voltage and amperage I to power it as long as it is running on 180w?
Yes, the voltage and amperage do make a difference.
To start with the motor wants to spin at a speed proportional to voltage, so if the voltage is too high or too low the motor will either over-speed or not run fast enough. Over-speeding can produce excessive brush arcing and commutator wear, reduce bearing life, or even cause the armature to fly apart.
Secondly, 'copper' loss (caused by resistance of the winding wires) is proportional to current squared, so running higher current at lower voltage results in higher loss at the same input power. This is only partially compensated by the lower 'iron' loss (caused by hysteresis and eddy currents induced into the magnetic core) at lower rpm. If the motor has a built-in cooling fan it may also run hotter due to less airflow at the lower speed.
If the battery voltage is higher than the rated motor voltage then you can reduce top rpm by limiting the throttle to less than 100%, but the speed controller must still be able to handle the full motor current (which with a PWM controller will be higher than the battery current).
Bottom line is you should find out what voltage the motor is designed to run on. One way might to run it on eg. 12V and measure the rpm, then relate that to what rpm it ran at in the go cart.
If you apply 180W to the motor when it is almost stalled (ie too much mechanical load), it will melt the motor.
If you apply 180W with no load, the motor will spin so fast that it will damage the device.
180W only applies to the rated torque and rated speed, which has about half of the no load speed and half the 0 rpm torque. I, too, was surprised when I learned that the rated values are for only one operating point of an infinite possible conditions.
Having said that, you could take a guess. Find the holding current and halve it. Then solve 180W = V * I, using this halved value of current, and that is your maximum voltage... The rated voltage. Then you may want to add a 25% factor of safety (ie apply less voltage).
But if you are using this for anything else than a hobby, do find out the true rated values.