Recall that work can be defined as the change in energy of a system. For most systems, we deal with two types of energy: potential and kinetic energy; therefore, work is defined as follows.

\(\displaystyle W = \Delta KE + \Delta PE\)

Where \(\displaystyle W\) is work, \(\displaystyle \Delta KE\) is change in kinetic energy, and \(\displaystyle \Delta PE\) is change in potential energy. Work can be negative or positive based on the type of work being performed. If work is done by an on object (such as object A) then it will have a positive value for work. However, if work is being done on the object (such as object B) then the object will have a negative value for work. We already defined work as being the sum of change in potential energy and change in kinetic energy; therefore, this sum will be positive for object A (positive work) and negative for object B (negative work). Note that both object A and object B will have the same, absolute value for work.

\(\displaystyle W=E_f-E_i\)

Initially there is only potential energy, and in the final state there is both kinetic and potential.

\(\displaystyle W=mgh_f+.5mv^2_f-mgh_i\)

Plug in all values and solve:

\(\displaystyle W=15000*9.8*59+.5*15000*2^2-15000*9.8*5\)

\(\displaystyle W=7.97*10^6J\)

Since the only force acting is gravity, we can determine work by determining the potential energy change. 

Since potential energy is given by:

\(\displaystyle PE=mgh\), where \(\displaystyle m\) is mass, \(\displaystyle g\) is the gravitational constant, and \(\displaystyle h\) is height above the earth, we can determine work done by determining:

\(\displaystyle W=mg(h_{final}-h_{initial})\)

We can assume \(\displaystyle h_{initial}=0\)

To determine \(\displaystyle h_{final}\), we determine the height above the ground that the object is moved. We do this by doing:

\(\displaystyle sin(60^o)=\frac{h}{10}\)

\(\displaystyle 5\sqrt{3}m=h\)

Therefore,

\(\displaystyle W=1kg*10\frac{m}{s^2}*5\sqrt(3)m=50\sqrt{3}J\)

We can use the expression for conservation of energy:

\(\displaystyle E = U_i + K_i + W = U_f + K_f\)

Since the rod is both initially and finally at rest, we can removed both kinetic energies. Also, if we assume point p is at a height of 0, we can removed initial potential energy, leaving us with:

\(\displaystyle W = U_f\)

Plugging in the expression for potential energy and expanding for both masses:

\(\displaystyle W = mgh = m_Agh_A + m_Bgh_B\)

Since the rod is vertical, we know that mass A is half a rod's length above our reference height, and mass B is half a rod's length below it. Thus we get:

\(\displaystyle W = m_Ag\left (\frac{1}{2}l \right )+m_Bg\left (-\frac{1}{2}l \right )\)

Factoring to clean up our expression:

\(\displaystyle W = \frac{1}{2}gl\left ( m_A-m_B\right )\)

We know all of our variables, so time to plug and chug:

\(\displaystyle W = \frac{1}{2}(10\frac{m}{s^2})(4m)(10kg-6kg)\)

\(\displaystyle W = 80J\)

The box is moving horizontally across the surface, which means that only the horizontal component of the applied force will do work on the box.  The horizontal component of the force is given by 

\(\displaystyle F_x= F_a \cos \theta\).  Therefore, to calculate the work done by the applied force, we will use the standard definition of work, given as

\(\displaystyle W=\vec{F}_a \cdot \Delta \vec{r} = F_x \Delta x = F_a \cos \theta \Delta x = 40N \cos \left( 20 ^{\circ} \right )4m=150J\)

In general,

\(\displaystyle W = \vec{F} \cdot \Delta \vec{x} = F \Delta x \cos \theta\)

In order to calculate the work done by a force, we need to find the angle between the force and the displacement through which the object moves.  Since the displacement is directed down the incline, then this means the friction force acts up the incline.  This means \(\displaystyle \theta = 180 ^{\circ}\).  We must remember the frictional force is represented as

\(\displaystyle F_f = \mu _k N\)

Where \(\displaystyle N\) is the normal force and \(\displaystyle \mu_k\) is the coefficient of kinetic friction.  On an inclined plane, we can show that \(\displaystyle N = mg \cos \theta\).  Therefore, we can finally write 

\(\displaystyle W_f = F_f \Delta x \cos(180^{\circ}) = -\mu _k mg \cos \theta \Delta x\) 

Plugging in everything, and noting the magnitude is the absolute value of a quantity, we can plug in for our answer:

\(\displaystyle \left| W_f \right|= (0.4)*90kg*9.8 \frac{m}{s^2}* \cos (50 ^{\circ})*15m=3401J \approx3400J\)

Because the normal force \(\displaystyle F_N\) is always perpendicular to the displacement \(\displaystyle \Delta \vec{r}\), the normal force does no work on the block, so \(\displaystyle W_N = 0J\)

This can also be shown mathematically by plugging in zero for theta in the equation for normal force:

\(\displaystyle W=F*cos(0) *d=0\)

From the Work-energy theorem,

\(\displaystyle W = \Delta KE\).  

From conservation of energy, we see that all of the potential energy contained in the spring will be transferred into kinetic energy of the ball, shown as 

\(\displaystyle \frac{1}{2}k \Delta x ^2= \frac{1}{2}m v^2 = KE_f\) 

Noting the initial kinetic energy of the ball is \(\displaystyle KE_i = 0J\), we can see that the work done by the spring will be equal to the initial potential energy of the spring.

\(\displaystyle W_s = \Delta KE = \frac{1}{2}mv^2= \frac{1}{2}k \Delta x ^2 = \frac{1}{2}30 \frac{N}{m}*(0.05m)^2= 37.5mJ\)

This can be a tricky question. You need to rely on the work-energy theorem, which states:

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

Since the elevator is at rest at both the beginning and end, the net work is 0; there is no net change in energy, and therefore no work.

This theorem can be confusing to some since it completely negates potential energy. However, let's think about the situation presented in the problem. A force is required to raise the elevator, meaning that energy is put into the system. However, since it comes back to rest, all of the energy that was put in has been removed by the force of gravity, resulting in a net of zero work.

Determining magnitude of final velocity:

\(\displaystyle |\overrightarrow{v}|=\sqrt{v_x^2+v_y^2}\)

Plugging in values:

\(\displaystyle |\overrightarrow{v}|=\sqrt{250^2+250^2}\)

\(\displaystyle |\overrightarrow{v}|=353.6\frac{m}{s}\)

Using definition of kinetic energy:

\(\displaystyle KE=.5mv^2\)

Plugging in values:

\(\displaystyle KE=.5*1560*353.6^2\)

\(\displaystyle KE=9.75*10^7J\)

Since there was no initial kinetic energy:

\(\displaystyle W=9.75*10^7J\)