(a)
The displacement occurs in horizontal direction, so to calculate work done, we need to determine the horizontal component of the net force on the block. The external force F acts by pulling (not pushing), this means the angle is 31 above horizontal out (i.e., lifting the block slightly), not in (which would push the block down slightly).
The net force is:
[tex]F_{net}=F\cos 31^\circ-F_{fric}\\F_{fric}=\mu(mg-F\sin 31^\circ)\implies\\F_{net}=F(\cos 31^\circ+\mu \sin 31^\circ)-\mu mg\\F_{net}=152.7N[/tex]
The work done:
[tex]W = F\cdot d = 152.7N \cdot 76.5m= 11682J[/tex]
The work done on the block is 11682 Joules
(b) Friction acts in the opposite of the line of motion. Its magnitude is determined by the normal force of the surface induced by vertical forces acting on the block - in our case: the gravity and the vertical component of the pulling force. The magnitude of the friction force in terms of the given variables is described in (a) as F_fric:
[tex]F_{fric}=\mu(mg-F\sin 31^\circ)=10.2N[/tex]
The friction force does negative work with respect to the work done by the external force:
[tex]W_{fric}=-F_{fric}\cdot d=-10.2N\cdot 76.5m = -779J[/tex]
The magnitude of the work done by friction is the absolute value of the work, i.e., 779 Joules
