A 500-kg elevator is pulled upward with a constant force of 5500 N for a distance of 50.0 m. What is the work done by the 5500 N force?
A) 2.75 × 10^5 J
B) -2.45 × 10^5 J
C) 3.00 × 10^4 J
D) -5.20 × 10^5 J
I'm split between A and B...
A) because w=Fd, 5500Nx50.0m= 2.75x10^5 J
B) because w=Fd, where F=ma, m=500kg, a=-9.8m/s^2, then that F multiplied by 50m, to give me -2.45x10^5J