Kinetic and Potential Energy

From MyMCAT

Contents 1 Introduction 2 Energy 3 Kinetic Energy 4 Kinetic Energy and the Work-Energy theorem 5 Potential Energy (Local/Constant Gravity) 6 Conservation of Energy: Total Energy 7 An Example of Total Energy in Practice

Introduction

In the previous section, Work Introduction, we introduced the concept of Work, now we will expand on this to discuss how work is done and how it can be used in conjunction with the concepts of energy to solve many different problem types.

Energy

Energy, in physics, is a measure of the ability to do work. It describes how much work can be performed by a force and it applies to almost all areas of the physical sciences. Here we will focus on the work that can be done by moving objects (kinetic energy), gravity (gravitational potential energy), and springs (elastic potential energy) but always remember that energy is more general than this and can also be applied to magnetic fields, electric fields, chemical cells, or any other system that can produce meaningful work.

All forms of energy, not just energy in the form of work (ie mechanical energy), is measured in the SI unit of Joules.

Perhaps the easiest way to understand the various forms of energy in motion is to consider a basic example. Consider a cyclist initially at rest. The cyclist will use chemical energy (in his body) to accelerate the bicycle to a given speed. In doing so, the energy in his body has been transformed in kinetic energy. This speed can be maintained without further work (except in reality where a little work will be required to overcome friction and air-resistance). Kinetic energy is the energy of motion, and as long as they remain in motion they will have kinetic energy.

Kinetic Energy

The two factors which determine how much kinetic energy a moving object has are its mass and velocity. Thus a heavy cyclist moving at the same speed as a lighter cyclist will have more kinetic energy.

Kinetic Energy and the Work-Energy theorem

Because of the relationship between energy and the fact that energy can be transfered between bodies, one can easily determine how much work has been done on an object by measuring the change in kinetic energy. Suppose, the cyclist slides to a halt because of friction on the road surface. How can we determine the work done if we aren't given the force that was applied? And to complicate this matter, the force that was applied was not uniform as it was a collision. In short, we cannot use our work formula alone but we can use our new concept of kinetic energy. Since we are given the mass and the velocity of the car, we can determine its initial kinetic energy. Once the cyclist comes to a complete stop its final kinetic energy must be zero (as it is not moving) and so the energy must have been transfered somewhere.

In general, we can say the work done by a net force equals the change in kinetic energy of the object, or

1. What is final velocity of a 5.0kg object moving at 15 m/s if a force of 50N opposes it while it moves across a distance of 6 meters?

	5 m/s
	10 m/s
	15 m/s
	20 m/s
→	We can calculate the work done by the force, F*d, and we know what the initial velocity and mass of the block is. We are looking for the change in velocity, and thus we can say we need to find the change in kinetic energy. We have all the variables we need to relate work and kinetic energy, except for the final velocity (or final kinetic energy). Therefore we can say, KEfinal - KEinitial = Work = F*d. Rearranging this we get:
→
→	Notice that we made the force negative as it is in the opposite direction of what we are considering positive movement (the force is opposing). Now let us solve for final KE.
→
→	We can now solve for the final velocity from our calculated final KE.
→
→	Therefore if we rearrange and solve for vf we get 10.3m/s.

Your score is 0 / 0

Potential Energy (Local/Constant Gravity)

Let us now return to the example of the cyclist moving along the road at a constant velocity. Now consider what happens when the cyclist encounters a hill just high enough such that the cyclist coasts up to the top and comes to a complete stop without any effort. In this case, the kinetic energy is now gone as the bike is stationary, but where did the energy go then? Going up hill takes energy because one must overcome the gravitational force. This new form of energy is called gravitational potential energy. Although the bicycle is no longer moving and it looks as though the kinetic energy it had is gone, the energy has not been destroyed, it is merely being stored in a different form and as soon as the cyclist begins to move back down the hill it will be converted back into kinetic energy (this is why we do not need to pedal as we go downhill on a bicycle, gravity in a sense "pulls" us as the stored potential energy is convered back into kinetic energy).

Potential energy is related to the gravitational field, the mass of the object, and the height at which the object is raised above the ground level as follows,

In determining the height, one can use any arbitrary assignment for the ground level. For instance, in the case of the bicycle, we could say that the base of the hill is the zero potential point and accordingly the top of the hill would have the most potential because it has the highest elevation from the zero point. Alternatively we could perhaps say that sea level was the real zero potential point, in which case the base of the hill would have some potential but the top of the hill would have more. In fact, any value can be chosen so long as the calculations are kept consistent (because it is the differences in height that we are concerned with).

1. What is the gravitational potential energy for a 4 kg object that is lifted 5 m?

	392 J
	196 J
	92 J
	46 J
→	When we are determining the change in potential energy in lifting an object, it is always simplest to assume the starting height is 0. That being said then, we can use PE=mgh where h is the height we lift it from zero, therefore we get (4 kg)(9.8 m/s</sup>2)(5 m) = 196 kg*m2/s2 = 196 J.

2. Which object would have more potential energy, a book of weight m on earth 1 meter above the ground or a book of weight 1/2 m on the moon held 3 meters above the ground? (assume the the gravitational field strength on the moon is 1/3 that of the earth)

	the book on earth
	the book on the moon
	they are equal
	the potential energies cannot be compared because they have different masses.
→	To answer this question, we should calculate the potential energy in both situations.
→	On the earth, we have U = m1gearthhearth = m1gearth( 1m ) = m1gearth
→	On the moon, we have U = m2gmoonmoon = (1/2)m1(1/3)gearth (3m) = (1/2)m1gearth.
→	Comparing the two then, we get that the moon is half what the earth would be.

Your score is 0 / 0

Conservation of Energy: Total Energy

The law of conservation of energy states that energy cannot be created or destroyed. A consequence of this law, the total amount of energy in an isolated system must remain constant. If we consider our previous examples of work, kinetic energy, and potential energy, it becomes obvious to see that these energies can be converted from one form to the other depending on the situation, however total energy is always constant.

Written mathematically we can say,

and thus if we are considering kinetic and potential energy together

Note, that the above formula does not considering friction. In the case of friction we will need an additional term to take into account the fact that friction or external force is doing work against us. The below formula represents the final and complete conservation of energy formula, pay special attention to the sign of Work. Notice the formula says minus Work (usually friction), this is because the Work term itself is negative (as it opposes the direction of potential and kinetic). All energies must sum up to remain constant during the process.

An Example of Total Energy in Practice

Consider a 1000kg truck traveling at 25m/s which then slams on the breaks and skids to a stop. Assuming the truck experiences a 8000N frictional force, what distance does the car travel? Questions such as these can easily be solved if we consider ALL the different energies involved. Initially we know,

PE = 0 J (the surface is flat so we can ignore any potential energy due to varying heights)

KE = 1/2mv_i² = 1/2(1000)(25)² = 312 5000 J

After the skid to a stop, we know the final conditions:

PE = 0 J (see above, still no change in height)

KE = 0 J (the truck is no longer moving and stationary objects do not have kinetic energy)

And finally, we know that work was done by friction. Note that the force of friction opposed the direction of movement and so the angle is 180 degrees apart.

W = Fdcos(180)

Or, W = (-8000)d

The total energy of the system initially is 312 5000 J. Afterwards there is no KE or PE, therefore ALL of the energy must have been used up due to the work of friction. And so we can say that the external work performed (by friction) equals the total energy of the system

U_i = 312 000 J = 0 J (KE) + 0 J (PE) - Wext

312 000 J = - (-8000)d

And so, if we solve for distance we get, 39.1m.