Using

Determining initial kinetic energy:

Combining equations

Converting to and plugging in values:

Determining final kinetic energy:

Combining equations

Converting to and plugging in values:

Plugging in values:

Work is a measure of force and displacement . Because C.J. did not move the phone at all, no work was done.

Since there is no friction between the mass and slope, there are only two relevant forces acting on the mass: gravity and tension. Furthermore, since the block is not in motion, we know that these forces are equal to each other. Therefore:

Substituting in an expression for the force of gravity, we get:

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

The net work done is equal to the change in kinetic energy. So we must find the kinetic energy at both the initial and final velocities and subtract.

Using

Using

and

All energy will be kinetic. Combining equations:

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 :

Work exerted on an object is equal to the dot product of the force and displacement vectors, or the product of the magnitudes of the vectors and the sin of the angle between them:

The work exerted on the wagon in this problem is thus:

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.

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

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.

For this question, we're told that an object is normally lifted a certain vertical distance. However, a ramp is to be added in order to make it easier to lift the object to the desired position, and we're asked to find out how much easier it will be by determining how the force required to lift it will change.

First, we must approach this problem from the perspective of energy. Specifically, we need to look at the change in gravitational potential energy. When the object is lifted a certain distance, its gravitational potential energy will increase according to the following expression:

Furthermore, it's important to realize that the change in mechanical potential energy only cares about the final and initial positions; it does NOT care about the path taken to get from initial to final. Therefore, the change in mechanical potential energy for lifting the object directly up is exactly the same as if the object were to be moved up a ramp to the same vertical location.

What's more is that we can realize the gravitational potential energy will be increasing as it is lifted, thus we need to put energy into this process by doing work. We can write the expression for work as follows:

Since the amount of work done on both processes is the same, we can set the two expressions equal to each other as follows:

By rearranging, we obtain:

The expression shown above tells us how the force needed to transfer the object via the ramp is different from the force needed to lift the object directly.

Thus, the force needed to move the object via the ramp is halved compared to the original force. Essentially, since we doubled the distance, we halved the force.