The box is moving at constant speed, so acceleration=0 and the equation of the forces acting on the box along the ramp becomes:
[tex]F-F_f-W_p = 0[/tex] (1)
where F=200 N is the force that pushes the box, Ff is the frictional force, and Wp is the component of the weight parallel to the ramp.
The frictional force can be written as
[tex]F_f = \mu m g cos \theta[/tex]
where [tex]\mu=0.18[/tex] is the coefficient of friction, m is the mass of the box, [tex]g=9.8 m/s^2[/tex] is the frictional force and [tex]\theta=30^{\circ}[/tex] is the angle of the ramp.
The component of the weight along the ramp is
[tex]W_p = mg sin \theta[/tex]
Substituting this into eq.(1), we have
[tex]F-\mu mg cos \theta- mg sin \theta=0[/tex]
and we can find the mass of the box:
[tex]m=\frac{F}{\mu g cos \theta + g sin \theta}=\frac{200 N}{(0.18)(9.8 m/s^2)(cos 30^{\circ})+(9.8 m/s^2)(sin 30^{\circ})}=31.1 kg[/tex]
Now we can also find the work done on the box. this is given by the gain in potential energy of the box, since there is no change in kinetic energy (the speed of the box is constant). Since the box has risen vertically by 1 m, the gain in potential energy (and the work done) is
[tex]W=\Delta U=mg \Delta h=(31.1 kg)(9.8 m/s^2)(1 m)=304.8 J[/tex]
So, the two answers are
(a) 304.8 J
(b) 31.1 kg
