physics
posted by Drew on .
Ice fishing equipment weighing 3,000 N is pulled at a constant speed V = 0.5 m/s across a frozen lake by means of a
horizontal rope. If the coefficient of kinetic friction is 0.05, what is the work done (in J) by the fisherman in pulling the
equipment a distance of 500 m?

Just an explanation without numbers would be awesome, i just want to understand the correct steps to take.
Thanks! 
W= (F(fr),s) = F(fr) •s•cosα.
α= 0, cosα= 1
F(fr)=μ•N=
W= μ•mg •s = 0.05•3000•500 = 
thanks, but how come m is not in kg?

weight= mg = 3000 N
m= weight/g=3000/9.8 =306.12 kg