If energy is conserved, then the total energy at the beginning equals the total energy at the end.

Since we have ONLY potential energy at the beginning and ONLY kinetic energy at the end, \(\displaystyle PE_i=KE_f\).

Therefore, since our \(\displaystyle PE_i=350J\), our kinetic energy will also equal \(\displaystyle 350J\).

The work-energy theorem states that work is equal to change in energy, or \(\displaystyle W=\Delta U\).

Total mechanical energy is the sum of potential and kinetic energies: \(\displaystyle \Delta U=\Delta PE+\Delta KE\)

In this case, she starts with \(\displaystyle 100J\) and ends up with \(\displaystyle 75J\). Since there was a change of \(\displaystyle -25J\), that means at some point during the system, \(\displaystyle 25J\) of work was done by the skier.

\(\displaystyle \Delta U=(PE_f-PE_i)+(KE_f-KE_i)\)

\(\displaystyle \Delta U=(0J-100J)+(75J-0J)\)

\(\displaystyle W=\Delta U=-25J\)

For this problem, we must use the law of conservation of energy.

\(\displaystyle PE_1+KE_1=PE_2+KE_2\)

\(\displaystyle mgh_1+\frac{1}{2}mv_1^2=mgh_2+\frac{1}{2}mv_2^2\)

Since the initial velocity is zero, there is no initial potential energy. Since the final height is zero, there is no final potential energy. This means that the final kinetic energy equals the initial potential energy.

\(\displaystyle PE_1=KE_2\)

\(\displaystyle mgh_1=\frac{1}{2}mv_2^2\)

The mass can be canceled from both sides.

\(\displaystyle gh=\frac{1}{2}v^2\)

Now we need to isolate the velocity by multiplying both sides by two, and taking the square root.

\(\displaystyle 2gh=v^2\)

\(\displaystyle \sqrt{2gh}=v\)

Remember the law of conservation of energy: the total energy at the beginning equals the total energy at the end. In this case, we have only potential energy at the beginning and only kinetic energy at the end. (The initial velocity is zero, and the final height is zero).

\(\displaystyle PE_i=KE_f\)

If we can find the potential energy, we can find the kinetic energy. The formula for potential energy is \(\displaystyle PE=mgh\).

Using our given values for the mass, height, and gravity, we can solve using multiplication. Note that the height becomes negative because the book is traveling in the downward direction.

\(\displaystyle PE=3.23kg*-9.8\frac{m}{s^2}*-3.01m\)

\(\displaystyle PE=95.28J\)

The kinetic energy will also equal \(\displaystyle 95.28J\), due to conservation of energy.

For this problem, use the law of conservation of energy. This states that the total energy before the fall will equal the total energy after the fall. The initial kinetic energy will be zero, and the final potential energy will be zero; thus, the initial non-zero potential energy will be equal to the final non-zero kinetic energy.

\(\displaystyle PE_i+KE_i=PE_f+KE_f\)

\(\displaystyle PE_i=KE_f\)

We can use the energy equations to define these equal energies:

\(\displaystyle PE_i=mgh_i\ \text{and}\ KE_f=\frac{1}{2}mv_f^2\)

The energies are equal, so we can say:

\(\displaystyle PE_i=KE_f=mgh\)

To solve this problem, use the law of conservation of energy. The skier initially starts at rest; all of his initial energy will be potential energy. At the bottom of the hill, the potential energy will be zero and all of the final energy will be kinetic energy. We can set these two values equal to one another based on the conservation of energy principle.

\(\displaystyle E_{initial}=E_{final}\)

\(\displaystyle PE_i=KE_f\)

Expand this equation to include the formulas for potential and kinetic energy.

\(\displaystyle mgh=\frac{1}{2}mv^2\)

Notice that the mass cancels out from both sides. This allows us to calculate without knowing the mass of the skier.

\(\displaystyle gh=\frac{1}{2}v^2\)

Plug in our given values for the height of the slope and acceleration due to gravity. Since potential energy is a state function (independent of the path) the slope of the hill is irrelevant.

\(\displaystyle 9.8\frac{m}{s^2}*11.7m=\frac{1}{2}v^2\)

\(\displaystyle 114.66\frac{m^2}{s^2}=\frac{1}{2}v^2\)

\(\displaystyle 229.32\frac{m^2}{s^2}=v^2\)

\(\displaystyle \sqrt{229.32\frac{m^2}{s^2}}=\sqrt{v^2}\)

\(\displaystyle 15.14\frac{m}{s}=v\)

For this we can consider the work-kinetic energy theorem.  As work is done on the object, its kinetic energy is changing.  In this case we have two different situations to consider.  In the first we must consider the horizontal force acting on the box alone.  In the second we must consider the horizontal force being resisted by a frictional force.

Let’s begin with the horizontal force acting alone.

Work is equal to the force times the displacement of the object.

\(\displaystyle W = F*d\)

In the first section the only force is \(\displaystyle 300N\) and the displacement is \(\displaystyle 10m\).

\(\displaystyle W = 300N*10m\)

\(\displaystyle W = 3000Nm\)

We can use the work kinetic energy theorem to solve for the change in kinetic energy during this first section

\(\displaystyle W = \Delta KE\)

\(\displaystyle W = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2\)

Since the initial velocity is zero the equation becomes

\(\displaystyle W = \frac{1}{2}mv_f^2\)

We can now plug in our values

\(\displaystyle 3000Nm = \frac{1}{2}*(100kg)*v_f^2\)

\(\displaystyle 3000Nm = 50kg*v_f^2\)

\(\displaystyle 60 = v_f^2\)

\(\displaystyle \sqrt{60} = v_f\)

\(\displaystyle 7.76m/s^2 = v_f\)

This is the velocity of the box after the first \(\displaystyle 10m\).  Now it is time to analyze the motion of the box when it has both friction and the applied force.

Newton’s 2nd law says that the net force is equal to the sum of the forces involved.

\(\displaystyle F_{net} = F_{app} - F_f\)

We need to find the friction force.

\(\displaystyle F_f = \mu F_N\)

The normal force in this case is equal to the force of gravity

\(\displaystyle F_N = F_g\)

\(\displaystyle F_g = mg\)

\(\displaystyle F_g = (100kg)(9.8m/s^2)\)

\(\displaystyle F_g = 980N\)

\(\displaystyle F_f = (0.25)(980N)\)

\(\displaystyle F_f = 245N\)

\(\displaystyle F_{net} = 300N - 245N\)

\(\displaystyle F_{net} = 55N\)

We can now determine the work on the box through the next \(\displaystyle 10m\).

\(\displaystyle W = Fd\)

\(\displaystyle W = (55N)(10m)\)

\(\displaystyle W = 550Nm\)

Like we did before we can now find the change of kinetic energy.  This time we will use the final kinetic energy from the first part as the initial kinetic energy of the second part.

\(\displaystyle W = \Delta KE\)

\(\displaystyle W = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2\)

\(\displaystyle 550Nm = \frac{1}{2}(100kg)v_f^2 - \frac{1}{2}(100kg)(7.76m/s)^2\)

\(\displaystyle 550Nm = (50kg)v_f^2 - 3010.88Nm\)

\(\displaystyle 3560Nm = (50kg)v_f^2\)

\(\displaystyle 71.22 = v_f^2\)

\(\displaystyle \sqrt{71.22} = v_f\)

\(\displaystyle 8.44m/s = v_f\)

Therefore the box will have a final velocity of \(\displaystyle 8.44m/s\).

We must consider several points during Mike’s jump off of the bridge.  The first point is when he is at the top of the bridge when he is about to jump.  The second point is the \(\displaystyle 12m\) below the bridge, just when the bungee cord would begin to stretch.  The third is the point at the bottom of the cord when it is fully stretched out.

To start let, us consider the first two points, when he jumps off the bridge and when he reaches \(\displaystyle 12m\) below the bridge.  For this first consideration, I will assume that our zero point of reference is \(\displaystyle 12m\) below the bridge.

At the top of the bridge, Mike has gravitational potential energy.  \(\displaystyle 12m\) later, all of this potential energy has been converted to kinetic energy.  According to the law of conservation of energy we can set these two things equal to each other.

\(\displaystyle GPE = KE\)

\(\displaystyle mgh = \frac{1}{2}mv^2\)

Since mass is in both sides of the equation it can be cancelled out to leave us with

\(\displaystyle gh = \frac{1}{2}v^2\)

We can now solve for the final velocity, just before the cord stretches.

\(\displaystyle (9.8m/s^2)(12m) = \frac{1}{2}v^2\)

\(\displaystyle 117.6 = \frac{1}{2}v^2\)

\(\displaystyle 235.2 = v^2\)

\(\displaystyle \sqrt{235.2} = v\)

\(\displaystyle 15.34m/s = v\)

Now let us consider two new points, the point at which the cord starts to stretch, and the point at the bottom when the entire cord is stretched out.  We will consider the lowest point as our zero point of reference in this case.

At the top, Mike has kinetic energy and gravitational potential energy as he is moving and above our reference point.  At the bottom all of this energy has converted to elastic potential energy.  According to the law of conservation of energy we can set these two things equal to each other.

\(\displaystyle GPE + KE = EPE\)

\(\displaystyle mgh + \frac{1}{2}mv^2 = \frac{1}{2}kx^2\)

The cord is going to stretch the same distance that Mike starts above the ground so we can exchange our x value for h so that everything is in similar terms.

\(\displaystyle mgh + \frac{1}{2}mv^2 = \frac{1}{2}kh^2\)

We can now put in our values and start to solve for h.  We will use our velocity from the first part as the velocity that Mike has.

\(\displaystyle (73kg)(9.8m/s^2)h + \frac{1}{2}(73kg)(15.34m/s)^2 = \frac{1}{2}(50N/m)h^2\)

\(\displaystyle 715.4h + 8589.02 = 25h^2\)

We are left with a quadratic equation.  So  we will need to get everything over to one side and use our quadratic formula to solve this problem.

\(\displaystyle 25h^2 - 715.4h - 8589.02 = 0\)

The quadratic formula is

\(\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

\(\displaystyle x=\frac{-(715.4)\pm\sqrt{(-715.4)^2-4(25)(8589.02)}}{2(25)}\)

The two answer we get for this is \(\displaystyle 37.72m\) and \(\displaystyle -9.11m\).  The reasonable answer is \(\displaystyle 37.72m\).  This is the distance the cord will stretch.

To find the total distance below the bridge we will need to add the amount that the cord stretched to the \(\displaystyle 12m\) it took to fall before the cord stretched.

\(\displaystyle 37.72m + 12m = 49.72m\)

Mike will stop \(\displaystyle 49.72m\) below the bridge.

We can use potential energy to solve. Remember, your height and your gravity need to have the same sign, as they are moving in the same direction (downward). Either make them both negative, or use an absolute value.

Using conservation of energy, we know that \(\displaystyle PE_{top}=KE_{bottom}\). This tells us that the potential energy at the top of the hill is all converted to kinetic energy at the bottom of the hill. We can substitute the equations for potential energy and kinetic energy.

\(\displaystyle mgh = \frac{1}{2}mv^2\)

The masses cancel out.

\(\displaystyle gh=\frac{1}{2}v^2\)

Plug in the values, and solve for the velocity.

\(\displaystyle (-9.8m/s^2)(-22m)=\frac{1}{2}v^2\)

\(\displaystyle 215.6m^2/s^2= \frac{1}{2}v^2\)

\(\displaystyle 431.2m^2/^s2 = v^2\)

\(\displaystyle \sqrt{431.2m^2/s^2} = v\)

\(\displaystyle 20.77m/s=v\)

We can use conservation of energy to consider the energy at the top of the incline and the bottom of the incline.  At the bottom of the incline the sled has some velocity.  At the top of the incline the sled has gravitational potential energy.  According to the law of conservation of energy these two values must be equal.

\(\displaystyle KE = GPE\)

\(\displaystyle \frac{1}{2}mv^2 = mgh\)

The mass cancels out of the equation.

\(\displaystyle \frac{1}{2}v^2 = gh\)

The angle does not matter in this case because it is a frictionless surface and all energy is conserved.

\(\displaystyle \frac{1}{2}v^2 = (9.8m/s^2)(1.4m)\)

\(\displaystyle \frac{1}{2}v^2 = 13.72\)

\(\displaystyle v^2 = 27.44\)

\(\displaystyle v = \sqrt{27.44}\)

\(\displaystyle v = 5.24m/s\)

The initial velocity of the sled is \(\displaystyle 5.24m/s\).