(a) 38.1 J
The work done by the applied force is given by:
[tex]W=F d cos \theta[/tex]
where
F = 20.0 N is the magnitude of the applied force
d = 2.10 m is the displacement of the block
[tex]\theta=25.0^{\circ}[/tex] is the angle between the direction of the force and the displacement of the block
Substituting into the formula, we find
[tex]W=(20.0 N)(2.10 m)(cos 25.0^{\circ})=38.1 J[/tex]
(b) 0
As before, the work done by the normal force exerted by the table is:
The work done by the applied force is given by:
[tex]W=F d cos \theta[/tex]
where
F is the magnitude of the normal force
d = 2.10 m is the displacement of the block
[tex]\theta=90.0^{\circ}[/tex] is the angle between the direction of the force and the displacement of the block, and it is 90 degrees because the block is moving along a horizontal surface, while the normal force is vertical
Since [tex]cos 90^{\circ}=0[/tex], therefore, the work done by the normal force is zero as well.
(c) 0
As in part c), the force (gravity) is perpendicular to the displacement (because gravity acts vertically, while the displacement of the block is horizontal), therefore the work done by gravity is zero as well.
(d) 38.1 J
Since the surface is frictionless, there is no friction acting on the block. Therefore, there are only 3 forces acting on the block: the applied force F, the gravity, and the normal force exerted by the table. Since the last 2 forces do zero work on the block, the total work done on the block is just the work done by the applied force, therefore 38.1 J.
