A 500 lb is pulled along the ground at constant speed for 15 ft,using a cable that makes 10owith the horizontal. Calculate the tension in the cable and the work done by this force. The coefficient of kinetic friction between ground and crate is mk = 0.3.

1) 144.67 lb, 7386.05 ft.lb.
2) 111.03 lb, 1234.89 ft.lb.
3) 456.23 lb, 2435.56 ft.lb.
4) 567.45 lb, 3245.45 ft.lb.

The horizontal components are,
Fx = 0 ; Tcos 10 - 0.3N = 0;

The vertical components are,
Fy = 0 ; N + Tsin 10 - 500 = 0;

The tension in the cable is,
T = 144.67lb.
The normal reaction of the surface is,
N = 124.88lb.


The work done in the process is,
Ur = 500(cos 10)(15)
= 7386.05ft.lb.