A husky exerts a constant horizontal pulling force of 160N over a distance of 15m on a loaded sledge of mass 100kg on snow. The sledge accelerates from rest to a speed of 4.0m/s. What is the work done by the husky against the friction on the sledge?
Name:
Anonymous2006-11-13 5:54
W = FD, so W of Husky = 2400 Newton Meters
Vf^2 = Vi^2 + 2AD
16 = 0 + 2A(15)
A = 16/30
F = MA, so W = MAD. Net work must be = (100)(16/30)(15), which = 800
Since the Husky did 2400 NM, but the net work was 800 NM, friction must've done -1600 NM. Somebody check my work, I've forgotten most of my physics already.