Calculate the minimum power output of an electric motor that lifts a 1.30 × 10^4-newton elevator car
vertically upward at a constant speed of 1.50 meters per second. [show all work, including
the equation and substitution with units.]

                 Work = (force) x (distance)

As the load is lifted, its gravitational potential energy
increases by

                       (1.3 x 10⁴ N) x (1.5 m) = 19,500 joules

every second. THAT's the energy that the motor must provide.

       19,500 joules/second  =  19,500 watts .

That's the minimum.  If there are frictional losses in the works,
gears, cables, or guts of the elevator, then the motor must provide
the extra energy required to overcome those too
.

Kinetic energy = 1/2mv^2. since the mass is constant we can form a ratio of it's kinetic energy at 60 divided by that at 30 to see ke(60)/ke(30) = +(60^2)/30^2). this may be said to 300% increase.