Ap physics, needed all work steps

Question

A 20 kg wooden crate is dragged 12 m across a rough level floor by a rope. The force of tension in the rope is 50 N at an angle of 25° above the horizontal and the crate is moving at a constant speed of 1.50 m/s.

1. Find the horizontal component of the force of tension in [N]. 
2. Find the vertical component of the force of tension in [N]. 
3. Find the magnitude of the weight of the crate in [N]. 
4. Find the magnitude of the normal force acting on the crate in [N]. 
5. Find the magnitude of the friction force acting on the crate in [N]. 
6. Find the work done by the tension force in [J].
7. Find the work done by the weight in [J].  
8. Find the work done by the normal force in [J].
9. Find the magnitude of the work done by the friction force in [J].  
10. What is the total work done on the object in [J]?
11. Find the coefficient of friction of the crate on the floor? 

Answer 1

Starting with the x and y components of tensional force:

F_Tx = F_T * cos(theta) = 50 * cos(25) = 45.3154 N

F_Ty = F_T * sin(theta) = 50 * sin(25) = 21.1309 N

Finding force of gravity F_g:

F_g = m*g = (20 kg) * (9.81 m/s^2) = 196.2 N

Solving for normal force F_N with zero acceleration:

Sum F_y = 0 = F_N + F_Ty - F_g

F_N = F_g - F_Ty = (196.2 N) - (21.1309 N)

F_N = 175.069 N

Solving for frictional force F_f with zero acceleration:

Sum F_x = 0 = F_Tx - F_f

F_f = F_Tx

F_f = 45.3154 N

Solving for work done:

W_T = F_Tx * D_x = (45.3154 N) * (12 m) = 543.785 J

W_g = F_g * D_y = (196.2 N) * (0 m) = 0 J

W_N = F_N * D_y = (175.069 N) * (0 m) = 0 J

W_f = F_f * D_x = (-45.3154 N) * (12 m) = -543.785 J

Finding coefficient of friction:

F_f = mu * F_N

mu = F_f / F_N = (45.3154 N) / (175.069 N)

mu = 0.258843