How much power must you exert to horizontally drag a 21.0kg table 15.0m across a brick floor in 20.0s at constant velocity, assuming the coefficient of kinetic friction between the table and floor is 0.500?

Answer 1

F = m*g = 21kg * 9.8N/kg = 205.8 N.

Fk = u*mg = 0.5 * 205.8 = 102.9 N.=Force
of kinetic friction.

Fe-Fk = m*a
Fe-102.9 = m*0 = 0
Fe = 102.9 N. = Force exerted.
Pe = Fe * d/t = 102.9 * 15/20 = 77.18 N.
= Power exerted.

Answer 2

Correction:
Pe=102.9 * 15/20 = 77.18 J/s=77.18 Watts
= Power exerted.