A 1,440-N crate is being pushed across a level floor at a constant speed by a force of 250 N at an angle of 20.0° below the horizontal;

(a) What is the coefficient of kinetic friction between the crate and the floor?
(b) If the 250-N force is instead pulling the block at an angle of 20.0° above the horizontal, as shown in the figure (b), what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in part (a).

Fc = 1440N @ 0 Deg. = Force of crate.

Fp = 1440*sin(0) = 0 = Force parallel to floor.
Fv = 1440*cos(0) = 1440 N. = Force perpendicular to floor.

Fv' = Fv + Fap*sin20.
Fv'=1440 + 250*sin20 = 1526 N=Normal.

a. Fn = Fap - Fp - u*Fv' = 0.
250*cos20 - 0 - u*1526 = 0.
1526u = 250*cos20 = 235.
u = 235 / 1526 = 0.154 = Coefficient of kinetic friction.

b. mg = 1440 N.
m = 1440/g = 1440 / 9.8 = 147 kg = Mass
of crate.

Fv' = 1440 - 250*sin20 = 1354 N. =[Normal.

Fn = Fap - Fp - u*Fv' = ma.
250*cos20 - 0 - 0.154*1354 = 147a.
235 - 209 = 147a.
147a = 26.
a = 26 / 147 = 0.177 m/s^2.

To solve these problems, we will use Newton's laws of motion and the concept of friction.

(a) To find the coefficient of kinetic friction between the crate and the floor, we need to consider the forces acting on the crate. The force of 250 N at an angle of 20.0° below the horizontal is the applied force. The force of friction opposes the motion and must be equal in magnitude and opposite in direction to the applied force to keep the crate moving at a constant speed. The weight of the crate (1,440 N) acts vertically downward.

Using trigonometry, we can decompose the applied force into its horizontal and vertical components:
F_horizontal = F_applied * cos(20.0°)
F_vertical = F_applied * sin(20.0°)

The force of friction can be calculated using the equation:
F_friction = coefficient_of_friction * F_normal

For a crate on a level floor, the normal force is equal in magnitude and opposite in direction to the weight of the crate:
F_normal = F_weight = 1,440 N

Since the crate is moving at a constant speed, the applied force is equal to the force of friction:
F_friction = F_horizontal

Therefore, we can write the equation:
F_horizontal = coefficient_of_friction * F_normal

Substituting the values, we have:
F_applied * cos(20.0°) = coefficient_of_friction * F_weight

Solving for the coefficient of kinetic friction:
coefficient_of_friction = (F_applied * cos(20.0°)) / F_weight

Substituting the given values:
coefficient_of_friction = (250 N * cos(20.0°)) / 1,440 N

Calculating this expression will give you the coefficient of kinetic friction between the crate and the floor.

(b) To determine the acceleration of the crate when the 250-N force is pulling at an angle of 20.0° above the horizontal, we need to consider the forces acting on the crate again. In this case, the applied force is now acting in the opposite direction to the crate's motion.

The force of friction still opposes the motion and has the same magnitude (coefficient_of_friction * F_normal). The weight of the crate (1,440 N) and the vertical component of the applied force (F_applied * sin(20.0°)) act vertically downward.

To find the net force acting on the crate, we need to subtract the forces acting in the opposite direction:
Net force = F_applied * sin(20.0°) - coefficient_of_friction * F_normal

Since the net force causes acceleration according to Newton's Second Law (F = ma), we can write:
Net force = mass * acceleration

The mass of the crate is given by the weight divided by the acceleration due to gravity:
mass = F_weight / g

Substituting the values, we have:
F_applied * sin(20.0°) - coefficient_of_friction * F_normal = (F_weight / g) * acceleration

Solving for the acceleration:
acceleration = (F_applied * sin(20.0°) - coefficient_of_friction * F_normal) / (F_weight / g)

Substituting the given values, including the coefficient of friction calculated in part (a), will provide you with the acceleration of the crate.