The work done is the product between the intensity of the force applied F, the amount of the displacement d of the book and the cosine of the angle [tex]\theta[/tex] between the direction of the force and the direction of the displacement:
[tex]W=Fd \cos \theta[/tex]
In our problem, the student is lifting the book, so he is applying a force directed upward, and the book is moving upward, so F and d are parallel and therefore the angle is zero, so [tex]\cos \theta = \cos 0=1[/tex]
Therefore, the work done is
[tex]W=Fd=(5 N)(2 m)=10 J[/tex]
