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:

\(\displaystyle W_{net} = \Delta U+\Delta K\)

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:

\(\displaystyle P = \frac{\Delta U}{t}=\frac{mg\Delta h}{t}\)

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

\(\displaystyle h = d\cdot sin (\theta)\)

\(\displaystyle P = \frac{mgd\cdot sin(\theta)}{t}\)

We can substitute velocity into this equation:

\(\displaystyle P = mgv\cdot sin(\theta)\)

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

\(\displaystyle P = (1500kg)(10\frac{m}{s^2})(20\frac{m}{s})\left sin(5^{\circ})\right = 26000W\)

Work is change in energy, so at a height of \(\displaystyle 5m\), the bag has more potential energy that when on the ground (zero gravitational potential energy). potential energy is equal to:

\(\displaystyle mgh=20kg\cdot 10\frac{m}{s^2}\cdot 5m = 1000J\)

Alternatively, \(\displaystyle Work = Fd\cos \Theta\).

\(\displaystyle \Theta\) is zero in this case, and \(\displaystyle F=mg=20kg\cdot 10\frac{m}{s^2}=200N\)

\(\displaystyle Work = 200N\cdot 5m = 1000J\)

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:

\(\displaystyle W=F\cdot d=Fdsin(\theta)\)

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

\(\displaystyle W=200N(5km)sin(35^{\circ})\)

\(\displaystyle W=573.6kJ\)

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

\(\displaystyle W_{outside}=\Delta PE+\Delta KE\)

 Here, \(\displaystyle \Delta PE\) is the change in potential energy, and \(\displaystyle \Delta KE\) is a change in kinetic energy. \(\displaystyle \Delta PE\) is \(\displaystyle 0\) because the object returns to the same height as when it was launched. \(\displaystyle \Delta KE\) however has changed because the object's velocity has changed.  Recall that the formula for the change in kinetic energy is given by:

\(\displaystyle \Delta KE= mv^2_{final}-mv^2_{initial}\)

Here \(\displaystyle m\) is the mass of the object, \(\displaystyle v_{final}\) is the final velocity of the object and \(\displaystyle v_{initial}\) is the initial velocity of the object. 

In our case:

\(\displaystyle m=10kg\), \(\displaystyle v_{final}=45 \frac{m}{s}\), and \(\displaystyle v_{initial}=50 \frac{m}{s}\)

\(\displaystyle \Delta KE= mv^2_{final}-mv^2_{initial}\)

\(\displaystyle \Delta KE=10kg*\left(27\frac{m}{s}\right)^2-10kg*\left(30\frac{m}{s}\right)^2\)

\(\displaystyle \Delta KE= 7290J-9000J= -1710J\)

\(\displaystyle W_{outside}=\Delta PE+\Delta KE\)

\(\displaystyle W_{outside}=-1710J\) 

Recall that work can be defined as:

\(\displaystyle W=F*d*cos(\theta)\)

Here, \(\displaystyle F\) is the magnitude of the force vector, \(\displaystyle d\) is the magnitude of the displacement vector, and \(\displaystyle \theta\) 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 \(\displaystyle 90^o\). Since \(\displaystyle cos(90^o)=0\), work done by the centripetal force on an object moving in a circle is always \(\displaystyle 0\).

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:

\(\displaystyle Work=3\ km(1000N)\)

\(\displaystyle Work=3000\ kJ\)

Work is a measure of force and displacement \(\displaystyle (W = F \cdot s)\). 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 \(\displaystyle 100N\) force is exerted. Using the equation

 \(\displaystyle Power = \frac{Force(distance)}{time}\) 

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

\(\displaystyle 640\:W= \frac{100N(distance)}{8.5 \:seconds}\)

\(\displaystyle distance = 54.4m\)

The answer is \(\displaystyle 0J\) 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:

\(\displaystyle W=f sin{\theta}d\)

Here, \(\displaystyle \theta\) is zero, so \(\displaystyle sin{\theta}\) is also zero, making the entire term, and thus the work zero.