Work and gravitational potental energy

[Shatzkinator]

Homework Statement

A thin rod, of length L = 2.00 m and negligible mass, that can pivot about one end to rotate in a vertical circle. A ball of mass m = 5.00 kg is attached to the other end. The rod is pulled aside to angle theta = 30 and released with initial velocity v= 0. As the ball descends to its lowest point, (a) how much work does the gravitational force do on it and b) what is the change in the gravitational potental energy of the ball - Earth system?.. etc.

Homework Equations

The Attempt at a Solution

In the solution, they show W = 5 x 9.8 x 2(1- cos30)

where do they get the 1-cos30??

 

[rootX]

W = Work at one step- Work at another step

= 5 x 9.8 x 2(1) - 5 x 9.8 x 2(cos30)

 

[nlightner]

The 1-cos30 comes from the fact that the gravitational force is not acting parallel to the displacement of the ball. This means that we need to take into account the component of the force that is acting in the direction of the displacement, which is given by cos30. So the total work done by the gravitational force is equal to the force (mg) multiplied by the displacement (2m) multiplied by the cosine of the angle between them (cos30), which gives us mg(2m)(cos30) or 5 x 9.8 x 2(1- cos30).