How much work must you do to push a 12 kg block of steel across a steel table at a steady speed of 1.3 m/s for 7.6 s ? the coefficient of kinetic friction for steel on steel is 0.60?

Work is force times distance. 
The distance is 1.3 m/s x 7.6 s = 9.88 m 

the force is only sufficient force to overcome friction. Assuming the table is a level table, the force to overcome friction is µ x normal force = 0.6 x (12 kg) x 9.8 m/s^2 = 70.56 N 

So the work is 70.56 N x 9.88 m = 697.13 J 

The power is simply the work / time = 697.13 J / 7.6 s = 91.7 or 92 Watts 

your answer is, orbit around the nucleus