Using the equation for conservation of energy, we get:

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:

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:

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 , therefore we can write:

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:

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

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

Rearranging for the angle, we get:

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

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

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

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

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:

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

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

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

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:

We substitute the known mass and solve for :

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

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

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:

Use conservation of energy:

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

Recall that the formula for gravitational potential energy is:

To determine the height on planet Mars needed to have equivalent potential energy, we can set the two equations for gravitational potential energies equal. 

, where  is mass and  is height above the ground on Earth while  is the height above the ground on Mars. 

The mass,  can be eliminated from both sides of the equation since they are equal. Plug in and solve for the height on Mars.