Sum the forces in the vertical axis:
#F_"net" = -mg darr +N_r uarr+P_vuarr#
were #mg# is weight, #N_r# is normal reaction and #P_v# is vertical component of the pulling force.
As the mass moves with constant velocity, there is no acceleration in the vertical direction, #F_"net" = 0#. We get
#N_r = mg - P_v = 10 xx9.81 - 50 sin25^@#
#N_r = 98.1 - 21.1=77N# ....(1)
Similarly in the horizontal axis:
#F_"net" = P_h(->) - F_f (larr)#
where #P_h# is horizontal component of the pulling force and #F_f# is force of friction.
As per given condition above there is no acceleration in the horizontal axis either, #F_"net" = 0#, and
#:.P_h = F_f#
Inserting given values ee get
#F_f=50cos 25^@=45.3N# ......(2)
If #mu_k# is coefficient of kinetic friction, then
#F_f = µ_kN_r#
Using (1) and (2)
#45.3= µ_kxx77#
#=>µ_k=45.3/77=0.59#, rounded to two decimal places.
