A232-kg crate is being pushed across a horizontal floor by a force p that makes an angle of 31.8 ° below the horizontal. the coefficient of kinetic friction is 0.269. what should be the magnitude of p, so that the net work done by it and the kinetic frictional force is zero?

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

grav force prop 1/r squared 2r all squared is 4rsquared 1/4rsquared quartered

spare some change and give it to the poor

explanation:

plz : ((