Suppose the crate is pushed a distance x. Only 2 forces do any work in the process:
• the horizontal component of p, with magnitude p cos(-31.8°) = p cos(31.8°)
• kinetic friction, with magnitude f = µn and where n is the magnitude of the normal force of the floor pushing upward on the crate and µ = 0.269.
If the net work done by these forces is 0, then
fx = p cos(31.8°) x ==> f = p cos(31.8°)
which in turn means the crate is pushed with constant speed.
By Newton's second law, we have
• net force acting on the crate in the horizontal:
∑ F = p cos(31.8°) - 0.269n = 0
• net force acting in the vertical:
∑ F = p sin(-31.8°) + n - mg = 0
Solve this system for p :
p cos(31.8°) - 0.269n = 0
-p sin(31.8°) + n = (232 kg) g = 2273.6 N
Multiply through the second equation by 0.269, then eliminate n :
p cos(31.8°) - 0.269n = 0
-0.269 p sin(31.8°) + 0.269n = 2273.6 N
==> p (cos(31.8°) - 0.269 sin(31.8°)) = 2273.6 N
==> p ≈ 3210 N
