This problem can either be very simple or very complex.

You could convert the contractor's potential energy into kinetic energy, and then find the velocity at which he hits the seesaw. Then, use that to calculate the kinetic energy of the bricks and the height at which the bricks will fly. Although some may find that fun to do, it's unnecessary.

If we assume that all of the energy of the man is transferred to the bricks, we can use the conservation of energy equation:

$\displaystyle E = U_i + K_i = U_f + K_f$

We need to make a few clarifying statements for our initial and final states. In the initial state, the contractor is at a height of 1.5m and not moving. The bricks are not moving at this point either. In the final state, the man is on the ground not moving, and the bricks are at a height of 10m and not moving. Therefore, we can rewrite the above equation as the following:

$\displaystyle m_cgh_i = m_bgh_f$

Rearranging for the mass of the bricks:

$\displaystyle m_b = m_c \frac{h_i}{h_f} = (80kg)(1.5m)(10m)= 12kg$

Using the equation for conservation of energy, we get:

$\displaystyle E = U_i+K_i=U_f+K_f$

Let's clarify that our initital state is when the ball is at its maximum height and the final state is when it reaches the ground. Plugging in our expressions and removing initial kinetic and final potential energy, we get:

$\displaystyle (0.85)(mgh_i) = \frac{1}{2}mv_f^2$

Note that we removed initial kinetic energy despite the ball moving. At the initial state, all of the ball's velocity is horizontal. Since there is no vertical velocity, we can ignore it for now, coming back to it later.

We multiply the initial energy by 0.85 because the problem statement says that we lose 15% of the ball's total energy after hitting the rim. Rearranging for final velocity:

$\displaystyle v_f = \sqrt{(0.85)(2gh_i)}$

$\displaystyle v_f = \sqrt{(0.85)(2\cdot10\frac{m}{s^2}\cdot5m)} = 9.22 \frac{m}{s^2}$

This is only the y-component of the final velocity. We need to combine this with the x-component to get the total velocity. The problem statement tells us that the final horizontal velocity is $\displaystyle 1\frac{m}{s}$, therefore we can write:

$\displaystyle v_f = \sqrt{v_{fy}^2+v_{fx}^2} = \sqrt{(9.22\frac{m}{s})^2+(1\frac{m}{s})^2}$

$\displaystyle v_f = 9.27 \frac{m}{s}$

This problem helps you become more comfortable with using obscure units. Since we are given work as a function of mass, we don't actually have to know the mass of the block. Furthermore, since we are neglecting air resistance, we know that all of the work is done by friction. To calculate the force of friction, we can use the following expression:

$\displaystyle F_f = \mu F_N = \mu mgcos(\theta)$

To calculate the work done by this force, we multiply by the distance that it was applied:

$\displaystyle W_f = F_fd = \mu mgdcos(\theta)$

Since we were given work as a function of mass, we can eliminate mass to get:

$\displaystyle \frac{W_f}{m}=\mu gdcos(\theta)$

$\displaystyle 10\frac{J}{kg}= \mu gdcos(\theta)$

Rearranging for the angle, we get:

$\displaystyle \theta = cos^{-1}\left (\frac{10\frac{J}{kg}}{\mu g d} \right )$

We know all of these values, allowing us to solve for the angle:

$\displaystyle \theta = cos^{-1}\left (\frac{10\frac{J}{kg}}{(0.70)(10\frac{m}{s^2}) (5m)} \right ) = 73^{\circ}$

We can use the expression for conservation of energy to solve this problem:

$\displaystyle E =U_i+K_i=U_f+K_f$

Assuming a final height of zero, we can eliminate final potential energy. Then, substituting in expressions for each variable, we get:

$\displaystyle mgh_i+\frac{1}{2}mv_i^2=\frac{1}{2}mv_f^2$

Canceling out mass and rearranging for final velocity, we get:

$\displaystyle v_f =\sqrt{2gh_i+v_i^2} = \sqrt{2(10\frac{m}{s^2})(3m)+(8\frac{m}{s})^2}$

$\displaystyle v_f = 11.1\frac{m}{s}$

The change in gravitational potential energy is only dependent on the final and initial height of the object, and is independent of the path taken to get to the final height. Because the change in height for the rock is zero, the change in gravitational potential energy equals zero.

Over the distance described, change in gravitational pull is negligible, so gravity can be treated as a constant. Potential energy therefore varies with the dumbell's proximity to the ground, h:

$\displaystyle PE=mgh$

Since the dumbell has an inital position of 100m, and an initial velocity of zero, its height can be described using the kinematic equation:

$\displaystyle h(t)=h_0-\frac{1}{2}gt^2$

$\displaystyle h(t)=100m-\frac{1}{2}(9.81\frac{m}{s^2})t^2$

$\displaystyle h(3s)=100m-\frac{1}{2}(9.81\frac{m}{s^2})(3s)^2$

$\displaystyle h(3s)=58.855m$

From this, the potential energy at a time of three seconds is:

$\displaystyle PE=20kg(9.81\frac{m}{s^2})(58.855m)$

$\displaystyle PE=11547J$

The first step for this problem is to determine the potential energy the ball has at the top of the ramp through this equation:

$\displaystyle U=m*g*h= (2kg)(10\frac{m}{s^2})(5m)= 100J$

We see that it has a potential energy of 100 joules. All of this potential energy gets converted to kinetic energy as the ball falls down the ramp. 

Knowing this, we can determine the velocity the ball has at the bottom of the ramp by setting the potential energy equal to the kinetic energy:

$\displaystyle 100J = \frac{1}{2}mv^2$

We substitute the known mass and solve for $\displaystyle v$:

$\displaystyle 100J=\frac{1}{2}(2kg)v^2$

$\displaystyle v^2= 100\frac{m^2}{s^2}$

$\displaystyle v=10\frac{m}{s}$

Gravitational potential energy is proportional to both the height and mass of an object. Gravitational potential energy is given by:

$\displaystyle PE_{grav}=mgh$ 

Gravitational potential energy is also relative to a "zero height" and in our case this is the ground. What this means is that if the ground is a our "zero height", the higher the object from the ground the greater the potential energy

$\displaystyle (2kg)(-9.8)(1.4m)=-27.4J$

Gravitational potential energy is proportional to both the height from a "zero potential energy" reference point (in our case the floor) and the mass. While the balls would have the same gravitational potential energy since they are the same mass, they have differing energies because of differing heights. The shortest cord, or the ball the highest from the ground will have the greatest potential energy as shown:

$\displaystyle PE_{grav}=mgh=(3kg)(-9.8)(1.5m)=-44.1J$

$\displaystyle PE_{grav}=mgh=(3kg)(-9.8)(1m)=-29.4J$

Use conservation of energy:

$\displaystyle E_{initial}=E_{{final}}$

$\displaystyle KE_1+PE_1=KE_2+PE_2$

$\displaystyle KE_1=PE_2$

$\displaystyle .5mv_i^2=mgh_f$

Plug in known values and solve for the maximum height of the object.

$\displaystyle \frac{.5v_i^2}{g}=h_f$

$\displaystyle \frac{.5*15^2}{9.8}=h_f$

$\displaystyle 11.5{m}=h_f$