The most important part of this question is noticing that it asks how much work has been done on the bar, not how much work the bodybuilder has exerted. Therefore we can use the work energy theorem:

Since the bar is initially at rest and returns to rest, the net work on the bar is zero. All of the energy exerted by the bodybuilder is counteracted by gravity.

Think about the system practically. Comparing the initial and final states, the bar is in the exact same position.

Since the truck is traveling at a constant rate, we know that all of the power exerted by the truck is going into a gain in potential energy. The power exerted will be a function of the change in potential energy over time. Therefore, we can write the following formula:

 is a vertical height, so we need to write that as a function of distance traveled up the slope:

We can substitute velocity into this equation:

We have values for all of these variables, allowing us to solve:

Work is change in energy, so at a height of , the bag has more potential energy that when on the ground (zero gravitational potential energy). potential energy is equal to:

Alternatively, .

is zero in this case, and

Work exerted on an object is equal to the dot product of the force and displacement vectors, or the product of the magnitudes of the vectors and the sin of the angle between them:

The work exerted on the wagon in this problem is thus:

Since this question refers to work done by nonconservative forces, we know that:

 Here,  is the change in potential energy, and  is a change in kinetic energy.  is  because the object returns to the same height as when it was launched.  however has changed because the object's velocity has changed.  Recall that the formula for the change in kinetic energy is given by:

Here  is the mass of the object,  is the final velocity of the object and  is the initial velocity of the object. 

In our case:

, , and 

Recall that work can be defined as:

Here,  is the magnitude of the force vector,  is the magnitude of the displacement vector, and  is the angle between the directions of the force and displacement vectors. In the case of circular motion, the force vector is normal to the circle since it points inward, and the displacement vector is tangent to the circle. This means that the angle between the force vector and displacement is . Since , work done by the centripetal force on an object moving in a circle is always .

Work is given by the dot product of force and displacement. Since both the force and displacement are in the same direction in this problem, work is simply the product of the two:

Work is a measure of force and displacement . Because C.J. did not move the phone at all, no work was done.

We can assume that the dog must carry his entire weight up the tree, and therefore a  force is exerted. Using the equation

we can use the evidence provided by the problem to solve for distance.

The answer is  because no work is done. For work to be done a force must be exerted across a distance parallel to the direction of the force. In this case, a force is exerted by the man but the wall is stationary and since it does not move there is no distance for work to take place on. The formula for work can be written as:

Here,  is zero, so  is also zero, making the entire term, and thus the work zero.