A) 0        B) 18 N     C) 21 N         D) 22 N           E) 44 N

three forces acting are (in the coordinate system shown)
a)  (0,N)  (-f_s,0)  (50sin25,-50cos25) SF = (50sin25- f_s)i + (N-50cos25)j = 0i + 0j       so   in y direction:    N-50cos25 = 0    N = 45 N  and    f_smax = m_sN = 0.50N = 0.50*45 = 22.5 N
in x direction:  50sin25- f_s = 0    so f_s = 50sin25 = 21 N very close to maximum but smaller so this is the answer.
b) (0,N)  (-f_k,0)  (50sin25,-50cos25)  SF = (50sin25- f_k)i + (N-50cos25)j   is for kinetic friction but we've already got above the correct result for static friction and we've found that the crate is at rest. Don't do the kinetic friction case.
 
 

2. The magnitude of the force (in newtons) required to cause an 0.04-kg object to move at 0.6 m/s in a circle of radius 1.0 m is:

A) 2.4 10^-2       B) 1.4 10^-2          C) 1.4 p 10^-2             D) 2.4p ²10^-2                   E) 3.13

 assuming that the T is the total force in radial direction (to cause an 0.04-kg object to move at 0.6 m/s in a circle of radius 1.0 m ) we have: T = mv²/r so T = 0.04 * 0.6² /1 = 0.0144 N = B
 
 
 

3.   A forward force of 3 lb. is used to pull a 60-lb sled at constant velocity on a frozen pond. The coefficient of friction is:
A) 0.5      B) 0.05       C) 2       D) 0.2         E) 20
     four forces acting are (in coordinate system shown) (0,N)   (T,0)   (0,-w)   (-f_k,0)
SF = (0+T+0-f_k)i +(N+0-w+0)j = m(0i + 0j)   ( constant velocity  means: acceleration is 0: means that acceleration vector is  0i + 0j)
in x direction:   T-f_k = 0  so:   3 - f_k = 0     so      f_k = 3
in y direction:   N-w = 0   so:     N = w (= 60)
f_k = m_kN    so:   3 = m_k60     and      m_k = 3/60 = 0.05