There will be no work done.

Energy is a scalar quantity. Since it has the same amount of energy in the end as the beginning, no work was done.

Alternatively, it can be modeled as the floor did negative work during the first half of the bounce (when the ball was compressing), and did positive work of the same magnitude and opposite sign in the second half of the bounce. This is due to the direction of the ball's motion flipping and the Force vector of the floor staying the same.

Recall that work can be defined as:

Here, is the magnitude of the force vector, is the magnitude of the displacement vector, and 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 . Since , work done by the centripetal force on an object moving in a circle is always .

If a person were to use their hand to push a book across the table, the type of force that their hand is applying to the book is classified as an electromagnetic force. Even though this may not seem intuitive, the reason is that the electrostatic repulsion of the atoms in the person's hand repel the atoms in the book. Due to this electrostatic repulsion between the atoms in the person's hand and the atoms in the book, the force is electromagnetic.

Gravitational forces occur due to gravity, that is, from an attraction between any two objects with mass. The strong and weak forces are both nuclear forces that act only across very, very short distances. Generally, these forces are only significant at the sub-atomic level in the atom's nucleus, and thus do not have a role at greater distances (such as between the person's hand and the book).

Using the equation of momentum,

we can compare the initial and final scenarios and set them equal to each other to solve for the final velocity of the block (don't forget to convert the units of to !):

Initially the car is at rest at the bottom of a hill, this velocity and height are zero.

Converting to

Plugging in values:

Use work:

All energy will be kinetic.

Convert to :

Plug in values. Force will be negative as it is directed against the direction of travel:

Solve for :

We can use the expression for conservation of energy:

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:

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

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:

Factoring to clean up our expression:

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

To solve this question, we need to recall the definition of work.

is work, is force, and is distance. Force is defined as:

To calculate work, we need to know about the distance travelled, mass, and acceleration. The question gives us the distance and mass but it doesn’t give us the acceleration of the object; therefore, we cannot calculate the work done by the person. Note that if the question gave us the force applied by the person, then we will be able to calculate the work without mass and acceleration.

Recall that work can be defined as:

Here, is the magnitude of the force vector, is the magnitude of the displacement vector, and 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 . Since , work done by the centripetal force on an object moving in a circle is always .

We begin by writing down the definition of an object's linear momentum

We then find the magnitude of the momentum by taking the square root of the sum of squares of its components.