# Energy The term "Energy" in itself is way, way too broad to condense into one file (I say this with a 2500 word note file, but you get what I mean). The term jumps around several topics; it can be used to describe the propensity for a spring to bounce back, or it can be allocated to a glass of hot water as a generalisation of its temperature... the possibilities are endless! A loose definition of energy is as follows: >Energy is the number we associate with many things, allowing us to categorise a change in the conditions we use to define the states. Energy exists all around us, from intrinsic quantum fields to the lightbulbs of your home. The symbiotic relationship between life and energy does cause problems, big and small, but that's another story for another time..? (blog maybe) I've a cool analogy - electoral votes. Energy can be transferred from a single source system to different systems, perhaps the Republicans have taken Idaho while the Democrats win over Wisconsin - but there can't be a loss in electoral votes, as there's a general maximum the energy source can give. Likewise, not every state can get all electoral votes - there's an efficiency rating on the heater that can get up to 100%, but that translates to a close election and many states won't like it. ...on second thought, maybe I'll stick to the textbook definition. >[!Success]- Units of Energy >You may have seen food manufactureres use "calories" to quantify energy. From a dietician's standpoint this would be acceptable, but we are physicists after all. In physics, the **Joule** is primarily used to describe energy. The definition of this unit is fluid and changes with context, so there is no rigorous definition for it! > >In case you were curious, one kilocalorie (kcal, unit they typically use) is equal to 4184 Joules. How's that to a three-course meal! # Work Done - A Brief Rundown ![[Pasted image 20231007000014.png]] When we think of "work done", we are typically using it to refer to energy that is being transferred in and out of an object. That is, if energy were being transferred to an object (perhaps to move it a certain distance), work would be positive, while energy being taken out of an object (such as to power a turbine) can be considered negative work. By accelerating an object, you are increasing its kinetic energy. Likewise, you must apply a force to decrease the kinetic energy of an object. These changes due to transfers in energy are known as **work done**. Remember that **conservation of energy** should apply unless the system is not closed. As a result, we can take a "transfer" of energy as just a transaction between two bank accounts - the balance on one account goes down while the balance on the other goes up. In general, the work done by a force is given by the formula: $W = Fd\cos \phi$ We need to find the **linear distance** travelled, so a triangle has to be created and trigonometry be used. Annoying, right? >[!Abstract]- More on deriving the Formula >Remember that we can write the total force on the object as: >$F=ma_{x}$ >We only refer to one dimension when we think of work done, so we can rewrite the velocity of the object as a function of the third SUVAT equation and the $x$ components of the object's motion. >$v^2 = u^2 + 2a_{x}d$ >Rearrange the equation to let us isolate $a_x$, and substituting into the first equation gives us: >$\frac{v^2-u^2}{2d} = a_{x}$ >$\frac{1}{2}mv^2 - \frac{1}{2}mu^2=F_{x}d$ >We can write $F_x$ as $F\cos x$, therefore giving us the equation in question. Naturally, there are constraints to this formula. Listed below are a couple of them, starting from: - The force must not change in magnitude or direction as the object moves - The object must act like a particle - meaning all parts of it MUST move together Mind you - the unit for work done is still the Joule! Remember that work done is just a measure of energy transfer between objects. Essentially, the change in the kinetic energy of an object can be interpreted as being the work done. This is defined rigidly as: $K_{f} = K_{i} + W$ Where $W = \Delta K$, $K_f$ is the final kinetic energy and $K_i$ is the initial kinetic energy. This is called the **work-kinetic energy theorem**. >[!Success]- Negative Work >Remember to take a look at the angle of which the force acts! For example, a force that acts 135 degrees against the direction of the force will lead to negative work, while a force that acts 45 degrees against the direction of the force will lead to positive work. There are also a few commonly-used presets when it comes to work done, such as finding the work done by a gravitational force or an elastic force. ## Case Study 1 - Work Done by a Gravitational Force %%diagram please%% For gravity, force equals to weight. Now substitute the equation for weight $mg$ to get: $W = mgd\cos \phi$ Given this equation we can therefore see that anything rising vertically up will lead to negative work as the kinetic energy of the object is being transferred into Gravitational Potential Energy. You can also relate the kinetic energy and the gravitational potential energy of an object through: $W_{k} = -W_{g}$ Where $W_k$ is the work done by the kinetic energy. Simply at that! ## Case Study 2 - Work Done by a Spring Force %%DIAGRAM PLEASE%% A spring force is a force created by a spring. Shocker, right? Springs themselves have a relaxed state - the base coil you are given, without any stretching or deformation. Springs act to restore themselves to this "relaxed state" if disturbed, which is the spring force in question. Let's start off by introducing **Hooke's Law**: $F_{x} = -kx$ Where $F_x$ is the force exerted on the spring, $k$ is the *spring constant*, an intrinsic property of all springs and $x$ is the extension of the spring in relation to the spring's relaxed state. We can imagine the spring as a point of mass on the origin of the x-axis: This means that we can measure displacement in relation to a negative-positive coordinate axis. For example, if you pulled a spring towards the right (so there would be positive displacement) the spring force would attempt to act against that force, leading to a leftward negative force. So to get the work done we've got a bunch of laws that we'll need to follow - such as treating the spring as always obeying Hooke's Law (most springs don't do this!) as well as treating it as massless. When we move a block attached to a spring, the spring will do work on the block reducing its kinetic energy. However, the value of $F_x$ is oten different as the block stretches the spring. meaning that at different instances the force is also different. This means we'll have to create an [[Calculus Basics (Maths)#The Integral|integral]] of all the individual segments within a spring to get the work done! $W = \sum -F_{x}x$ $W = -k\int_{x_{i}}^{x_{f}} x \, dx $ $W = -\frac{1}{2}kx_{i}^2 - \frac{1}{2}kx_{f}^2$ Where $x_i$ is the initial displacement and $x_f$ is the final displacement. Now an alternative to this would be: $W = \frac{F^2}{2S}$ Not the most fun thing to use. See why we do what we do? ## Final Boss - Work Done by **ANY** Force In general, forces have three dimensions. This means we're going to have to find the linear distance travelled on three different dimensions. Like the spring force, we can't use the formula $W = Fd\cos\phi$ because the force is variable, so calculus must be used. We can create an integral after plotting the force in relation to the displacement of the particle: $W = \lim_{ \Delta x \to 0 } \sum F_{j_{avg}} \Delta x$ $\int_{x_{i}}^{x_{j}} F(x)\, dx $ Where $x_i$ is the starting position, $x_j$ is the ending position of the particle having work done on it and $F_{j_{avg}}$ is the average position within the Riemann integral. If this force were three-dimensional we would have to create an incremental displacement vector $\overrightarrow{r}$, consisting of: $d\overrightarrow{r} = dx \hat{i} + dy \hat{j} + dz \hat{k}$ As a result we can write the work done as: $W = F \cdot dr = F_{x} \cdot d \hat{i} + F_{y} \cdot d \hat{j} + F_{z} \cdot d \hat{z}$ Given that we can express the rates of change for all of the variables, a simple integration with limits $x_i$ and $x_j$ can allow us to get an expression for $W$: $W = \int_{x_{i}}^{x_{j}} F_{x} \, dx + \int_{x_{i}}^{x_{j}} F_{y} \, dy + \int_{x_{i}}^{x_{j}} F_{z} \, dz$ This allows us to account for all the components of the force. Check out our [[Calculus Basics (Maths)#The Integral|integration document]] for more info on it! # Kinetic Energy This is just the energy that we associate with objects moving. It allows us to monitor the energy of a particle, and allows us to quantify the state of a particle moving. The formula of kinetic energy is given as: $K = \frac{1}{2}mv^2$ Simple as that! For work done, remember - any change in the kinetic energy will equal to the work done. $W = \frac{1}{2}m_{f}v_{f}^2 - \frac{1}{2}m_{i}v_{i}^2 = \Delta K$ # All About Potential Energy Another form of energy you'll have to look out for will be **potential energy**. We usually associate this form of energy with the internal arrangement of particles within an object, such as elastic potential energy or gravitational potential energy. To create a generalised formula for potential energy, let's take another look at the work kinetic-energy theorem. Remember that the kinetic energy of an object increases as the gravitational potential energy decreases. We can write this relationship as shown: $\Delta U = -W $ These systems always consist of two elements, and there are forces acting between the two systems. We can group the different types of forces that act on objects into two categories: **conservative** forces and **non-conservative** forces. So, now that we have that equation under our belt, we can rearrange the equation to let us isolate $W$. We can take the force as perfectly conservative. $W = \int_{x_{i}}^{x_{f}} F(x) \, dx $ Where $x_i$ and $x_{f}$ denote the upper and lower bounds for the work done we are trying to parse. Since the change in potential energy is the negative of the work done the equation for potential energy can thus be given as: $\Delta U = -\int_{x_{i}}^{x_{f}} F(x) \, dx$ This F(x) is a function of force against displacement (see above!) , so we can allow $x_i$ and $x_f$ to be the starting and final displacements of the object that is being moved. >[!Danger]- Extra Intuition for the formula for Potential Energy >We can take an example for this - if you stretch a spring, it's noted as having a large potential energy. The potential energy actually describes the **potential work done** by a system if released, so we can therefore model an increase in the work done by a system as a linear decrease in the potential energy (as the energy is being converted into a "real" form). There is one last formula we need to know for this - the **work-kinetic energy theorem**. Let's walk ourselves through on how to derive this! ## Example - The Gravitational Potential Energy of a Spherical Distribution Now, imagine that we've got a star (in [[Hydrostatics (Mechanics)#The Hydrostatic Equation|hydrostatic equilibrium]], of course), and we're going to have to find the gravitational potential of a region within it. ![[spherical shell.png]] Now, if we harken to the [[Gravitational Potential Energy]] doc, we'll know that the formula for the gravitational potential energy of an orbiting object is: $U = -\frac{GMm}{r}$ *Where, if you're unfamiliar with gravitation:* - $M$ is the **mass of the star** - $G$ is the **Universal Gravitational Constant** - $m$ is the **mass of the orbiting object** - $r$ is the **distance from the centre of mass** (the **barycenter**) Unfortunately, if the star's at the centre of the system, we can't really say that the sphere is 'orbiting' around it. That's why we'll need to take the gravitational potential energy at EVERY POINT in the sphere and add them up to find the agglomerate potential energy of the shell! For the spherical shell above, this gives: $U = \int_{0}^r \frac{GM}{r} \Sigma N\, dr $ Where we integrate along the full radius for this. This essentially means we're following a line along the radius of the shell, as you can see: ![[spherical shell with diagram.png]] To account for ALL the particles in the distribution, we can pretend that each infinitesimal point along $r$ will give a 'shell' of surface area $4 \pi r^2$, with a particle density $\rho(r)$. If we let the distance between each 'particle' tend towards 0, the space between each shell becomes negligible ![[spherical shell with distance.png]] *Two shells intersect along the guiding axis $r$ and have a distance $d$ between them. What type of approximation do you think we can make when $d$ tends towards the width of 1 particle?* Applying this logic in the context of counting particles will give us: $U = -\int_{0}^r \frac{GM(r)}{r} \rho \ 4\pi r^2\, dr $ If you're a maths purist, the $4\pi r^2$ term also comes from integrating through the volume to find the area within it, which is proportional to the number of particles within the spherical shell. The constant of proportionality between the integrated volume and the number of particles is the particle density, $\rho$. So, let's write everything in terms of $R$ first: $U = -\int_{0}^r \frac{G (\frac{4}{3} \pi r^3 \rho(r) ) \rho(r) \ 4\pi r^2}{r} \, dr $ $U = -G\int_{0}^r \frac{16}{3} \pi^2 r^4 \rho^2 \, dr $ $U = -G [\frac{16}{15} \pi^2 r^5 \rho^2]_{0}^r $ Remember that $M = \frac{4}{3}\pi r^3\rho$, where $M^2$ therefore equals to $\frac{16}{9} \pi^2 \rho^2 r^6$. So, let's substitute $M^2$ into our expression: $U = -\frac{16}{15}\pi^2r^5\rho^2 = -\frac{\frac{16}{15}GM^2}{\frac{16}{9}r}$ $U = -\frac{3GM^2}{5r}$ Done! To standardise this for ALL fields, the PE (potential energy) will just equal to the sum of the potentials of all the particles enclosed in the object. Note that the way to find the sum of all particles will be different though. It's important you get the hang of this - synergises VERY well with the [[The Virial Theorem|Virial Theorem]]. >[!Abstract]- The formula for the cylinder? %%it's all about that density! remember - DENSITY, IN THIS CASE, IS HOMOGENEOUS ACROSS THE STAR AND IS THUS NOT A FUNCTION OF RHO!%% ## The Concept of "Infinity" Remember that the force on an object is given by Newton's Second Law: $F = ma$ The equation $W = \int_{x_{i}}^{x_{f}} F(x) \, dx$ shows us that there is a relationship between the force (when one-dimensional) and the work done. Newton's Second Law therefore shows us that an object accelerates given a force is exerted onto it. %%link it back to the work-energy KE theorem!%% ## More on Conservative Forces One test we can use to determine the conservative forces in a system are to make an object travel around a **closed path** - that is, an ideal situation where a particle's net displacement is zero, such as a circle. If the total energy transferred to and from the object is equal to zero, the force can be considered conservative. >[!Success]- What comes up, must come down! >This phrase ties into our conservative forces - if shot in a perfect vacuum, a bullet travelling straight up would only have the gravitational pull of the Earth affecting it in any way. When the bullet inevitably falls down, it hits the Earth with the same speed as it did upon being launched, therefore containing the same kinetic energy from before. This is therefore a conservative force! To write this rigorously, we can imagine two points $a$ and $b$ which form a closed path. This means that a conservative force will satisfy: $W_{ab} = W_{ba}$ Where $W$ is the work done. Remember, it's energy, a scalar! # Mechanical Energy - Conservation of Energy >[!Warning]- CoE FAQ >This is a prerequisite for thermodynamics. Most thermal systems at this level will either revolve around using Lagrangians or the laws of Thermodynmaics, both of which we'll go through. > >For more about Lagrangians, finish everything non-multivariable about calculus before taking a look! I've nested a document in the "Additional" section if you'd like to take a look (not finished yet) We take the mechanical energy of *any* system as the sum of the kinetic and potential energies. $E_{mec} = K + U$ Remember that the change in kinetic energy and potential energy are directly proportional, just with opposite signs. We can visualise this through the relationship: $\Delta K = -\Delta U$ So if one energy decreases the other will increase with an equal magnitude. This means that the total energy in a system will stay the same: $K_{i} + U_{i} = K_{f} + U_{f}$ In other words, energy is conserved - hence conservation of energy. This can only happen in isolated (no external influences) systems where only conservative forces act. This ties into those forces - "**conservative**" forces get the name from satisfying conservation of energy! %%julien how could you have forgotten this? add it as the sum of the potential and the kinetic energies of a particle (in an orbit for example -- build to gravitation!)%% Sometimes, not even physicists want to be restrained by the backlog of potential and kinetic energies - instead opting to combine them together, forming the amalgamation known as "mechanical energy". ## Potential Energy Curves Sometimes, scientists want to get potential energy curves of objects, perhaps to allow them to relate back to the work done by the object. You'll never know where that'll be useful! Remember that the work done by a force is given by the force function, $F(x)$, and the change in distance travelled by the particle $\Delta x$. We can therefore write the change in potential energy as a function of both: $\Delta U(x) = -F(x)\Delta x$ We can isolate $F(x)$ and pass it through the differential limit (allowing $\Delta x$ to approach 0) giving us: $F(x) = -\frac{dU(x)}{dt}$ When dealing with strictly potential energy, this is given with an integral: $\Delta U(x) = -\int_{\infty}^{rel} F(x) \, dx $ Where the potential energy must be brought from an arbitrary infinite distance to the relative point to which it begins to fluctuate. So let's take a look at a couple potential energy graphs! One final thing - we can rewrite the conservation of energy equation as a component of kinetic energy and potential energy functions: $K(x) = E_{mec} - U(x)$ ![[potential energy curve.png]] As the caption says, the formula above decrees that the kinetic energy is going to equal to the mechanical energy of the system minus the potential energy $U$. The brown line is the total mechanical energy of the system. This means we can measure the speed of an object through taking kinetic energy measurements! We can also refer to this graph by calling it to be "in equilibrium" at certain points. When a potential energy curve meets the line of mechanical energy, the object does not move - this is a **turning point**. If this happens for a prolonged period of time where no force acts on the particle, we call this **neutral equilibrium**. However, if the mechanical energy were at, say, 3, the turning point would be at $x_3$ - which we call an **unstable equilibrium**, as the particle only has to be slightly displaced for it to begin moving. ## Work Done By External Forces Typically systems aren't perfect - typical redundancies suffered by systems include energy lost as heat thanks to friction. However, this friction still ties into the system, as the term "work done" does just mean a transfer of energy. Think of the bank account of energy as having proxies now! When you pull a bowling ball up to bowl it (unless you overhand throw), you apply a force to the bowling ball which results in work being done. This force leads to an increase in the mechanical energy, increases the kinetic energy of the bowling ball. and the GPE (Gravitational Potential Energy) of the bowling ball (assuming your bowling ball travels some distance upwards!) We can express this relationship as: $W = \Delta U + \Delta K $ >[!Abstract]- Deriving the Equation >When the bowling ball in the example above rolls on the floor, there is a kinetic frictional force which acts on the ball. This means that we can model the force on the bowling ball using Newton's Second Law: >$F - f_{k} = ma$ >Which gives us an equation for the resultant force. Since this acceleration is constant, we can use the third SUVAT equation $v^2 = u^2 + 2as$ and rearrange it to isolate $a$, giving us: >$a = \frac{v^2-u^2}{2s}$ >Substitute into the Newton's Second Law equation above: >$F-f_{k} = \frac{m(v^2-u^2)}{2s}$ >$Fs = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 + f_{k}s$ >We can generalise $\frac{1}{2}mv^2 - \frac{1}{2}mu^2$ into $\Delta K$: >$Fs = \Delta K + f_{k}s$ >The potential energy might change if the bowling ball were to be rolling down a ramp or if the internal heat of the ball was increasing, giving us: >$Fs = \Delta E_{mec} + f_{k}s$ >The change in thermal energy (or heat) of the ball is equal to the product $f_ks$ giving us a final equation for the work done by an external force: >$Fs = \Delta E_{mec} + \Delta E_{th}$ >Remember the work done equation? It came full circle! >$W = \Delta E_{mec} + \Delta E_{th}$ # Power Power is defined to be the **rate of the work done by a force**. In other words, we can give it with the formula: $P = \frac{W}{\Delta t}$ Mind you, this only gives the average power during a certain time. Sometimes the power changes (such as when people adjust the temperature on the shower knob). If we want to find the instantaneous power, a simple derivative should do the trick: $P = \frac{dW}{dt}$ >[!Tip]- Units >Power itself has several units that have different applications. We can first start with the **Watt**, given by the Newton-meter $mN$. >The **Kilowatt-Hour**, kWh, is also commonly used for larger power consumption readings, given by a denomination of $3.6 \times 10^6$ watts. We can also express power in terms of the work done as shown: $P = \frac{{F\cos \phi dx}}{dt} = F\cos \phi\left( \frac{dx}{dt} \right)$ But $\frac{dx}{dt}$ is just the magnitude of the velocity $v$, so we can rewrite the equation for instantaneous power: $P = Fv\cos \phi$ Remember that work done is given in Joules. This is the same unit as energy, meaning that the work done by an object is equal to the change in energy of an object. This means we can just substitute the equation for power with: $P = \frac{dE}{dt}$ Where $E$ is the energy of the system. # Extended - Potential Wells *This ties directly into Quantum Mech!* Similar to Potential Energy Graphs (see above), can also be used to define the *potential* of an object in a field. As per usual, we define the potential with the total energy of the particle, but instead defined with %%provisional explanation, don't upload - need to confirm%% ![[Pasted image 20240715162709.png]] Usually, we're given an arbitrary distance for the size of the 'potential well' - which is the distance $d$ in the diagram above. Aside from general classical mechanical uses, the potential well also sets the scene for quantum mechanics, forming part of the negative exponential slope that seems to be introductory QM. **TBA APPLICATIONS, NEED TO FINISH SOMETHING ELSE** >[!Abstract]- Applying the Quantum Well - Quantum Mechanics # Problems ## Work Done - Problems >[!Example]- Module 7-1 Problem 5 >A father racing his son has half the kinetic energy of the son, who has half the mass of the father.The father speeds up by 1.0 m/s and then has the same kinetic energy as the son. What are the original speeds of (a) the father and (b) the son? >**Solution** >a) We can make a simultaneous equation to express the relationship between father and son (if only!) >$\frac{1}{2}2m(v+1)_{1}^2 = \frac{1}{2}mv_{2}^2$ >$2mv_{1}^2 = \frac{1}{2}mv_{2}^2$ >Upon reducing we get: >$\frac{1}{2}2m(v+1)_{1}^2 = 2mv_{1}^2$ >$2m(v+1)_{1}^2 = 4mv_{1}^2$ >This implies that: >$v_{1}^2+2v_{1}+1 = 2v_{1}^2$ >$0 = v_{1}^2 - 2v - 1 = 2.41 m /s$ >So the dad's moving at the (very specific, need to ask for help) speed of 2.41m/s. We can substitute this value into the first equation and divide both sides by $m$ to get a value for the speed of the son: >$\frac{1}{2}2(3.41)_{1}^2 = \frac{1}{2}v_{2}^2 = 4.82 m /s$ >[!Example]- Module 7-2 Problem 9 >The only force acting on a 2.0 kg canister that is moving in an xy plane has a magnitude of 5.0 N. The canister initially has a velocity of 4.0 m/s in the positive x direction and some time later has a velocity of 6.0 m/s in the positive y direction. How much work is done on the canister by the 5.0 N force during this time? >**Solution:** >The question gives us the initial and final velocities of the object. This allows us to get an expression using an intermediary step for the work done equation, giving us: >$W = \frac{1}{2}mv^2 - \frac{1}{2}mv^2_{0} = \frac{1}{2}2(6^2) - \frac{1}{2}2(4^2)$ >This gives us a numerical value: >$36-16 = 20J$ >[!Example]- Module 7-3 Problem 17 >A helicopter lifts a 72 kg astronaut 15 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/10. How much work is done on the astronaut by >(a) the force from the helicopter and >(b) the gravitational force on her? >Just before she reaches the helicopter, what are her >(c) kinetic energy and >(d) speed? > >**Solution a)** > >Word salad! Note that the work done on the person by the helicopter is just equal to the gravitational force plus the extra $\frac{g}{10}$ of acceleration she has, giving us: >$W = F\times d = 72 \times (\frac{9.81}{10} + 9.81) \times 15 = 11654 J$ >**Solution b)** > >Remember that the work-kinetic energy theorem dictates that the work done is equal to the final minus the initial kinetic energy of the object. As the object has an initial kinetic energy of 0 joules, this gives us: >$15 \times 9.81 \times 72 = W$ >$W = 10594J \approx 10600 J$ > >**Solution c)** >Subtract the two joule values by each other (to get the total kinetic energy of the person) then substitute it into the kinetic energy equation: >$E_{k} = 11654 - 10600 = 1054$ >**Solution d)** > >Now substitute the numbers into the kinetic energy equation! >$1054 = \frac{1}{2}72 \times v^2$ >$v^2 = 29.3$ >$v = 5.41 m/s$ >[!Example]- Module 7-4 Problem 31 >The only force acting on a 2.0 kg body as it moves along a positive x axis has an x component $F_x = -6x N$, with x in meters.The velocity at x = 3.0 m is 8.0 m/s. (a) What is the velocity of the body at x = 4.0 m? (b) At what positive value of x will the body have a velocity of 5.0 m/s? >[!Example]- Module 7-5 Problem 36 >A 5.0 kg block moves in a straight line on a horizontal friction-less surface under the influence of a force that varies with position as shown in Fig. 7-39.The scale of the figure’s vertical axis is set by Fs = 10.0 N. How much work is done by the force as the block moves from the origin to x = 8.0 m? >[!Example]- Module 7-5 Problem 40 >A can of sardines is made to move along an x axis from x = 0.25 m to x = 1.25 m by a force with a magnitude given by F 􀀁 exp(􀀃4x2), with x in meters and F in newtons. (Here exp is the ex-ponential function.) How much work is done on the can by the force? >[!Example]- Module 7-6 Problem 46 >The loaded cab of an elevator has a mass of 3.0 􀀅 103 kg and moves 210 m up the shaft in 23 s at constant speed. At what aver-age rate does the force from the cable do work on the cab? >[!Example]- Module 7-6 Problem 49 >A fully loaded, slow-moving freight elevator has a cab with a total mass of 1200 kg, which is required to travel upward 54 m in 3.0 min, starting and ending at rest. The elevator’s counter-weight has a mass of only 950 kg, and so the elevator motor must help. What average power is required of the force the motor exerts on the cab via the cable? >[!Example]- Module 7-6 Problem 52 >A funny car accelerates from rest through a measured track distance in time T with the engine operating at a constant power P. If the track crew can increase the engine power by a differential amount dP, what is the change in the time required for the run? ## Potential Energy - Problems >[!Example]- Module 8-1 Problem 1 >What is the spring constant of a spring that stores 25 J of elastic potential energy when compressed by 7.5 cm? >**Solution:** >Substitute the potential energy $U = \frac{1}{2}kx^2$ with the corresponding values: >$25 = \frac{1}{2} k (56.25)$ >$25 = k (28.125)$ >$k = 1.125$ >[!Example]- Module 8-1 Problem 8 >A 1.50 kg snowball is fired from a cliff 12.5 m high. The snowball’s initial velocity is 14.0 m/s, directed 41.0N above the horizontal.a) How much work is done on the snowball by the gravitational force during its flight to the flat ground below the cliff? (b) What is the change in the gravitational potential energy of the snowball – Earth system during the flight? (c) If that gravitational potential energy is taken to be zero at the height of the cliff, what is its value when the snowball reaches the ground? >[!Example]- Module 8-2 Problem 20 >A pendulum consists of a 2.0 kg stone swinging on a 4.0 m string of negligible mass. The stone has a speed of 8.0 m/s when it passes its lowest point. > >(a) What is the speed when the string is at 60$^\circ$ to the vertical? > >(b) What is the greatest angle with the vertical that the string will reach during the stone’s motion? > >(c) If the potential energy of the pendulum – Earth system is taken to be zero at the stone’s lowest point, what is the total mechanical energy of the system?* > >**Solution:** >Let's start off with a diagram! >[!Example]- Module 8-3 Problem 38 >Figure 8-49 shows a plot of potential energy U versus posi-tion x of a 0.200 kg particle that can travel only along an x axis under the influence of a conservative force. The graph has these values: UA = 9.00 J, UC = 20.00 J, and UD = 24.00 J. The particle is released at the point where U forms a “potential hill” of “height” UB = 12.00 J, with kinetic energy 4.00 J. What is the speed of the particle at (a) x = 3.5 m and (b) x = 6.5 m? What is the position of the turning point on (c) the right side and (d) the left side? >[!Example]- Module 8-4 Problem 43 >A collie drags its bed box across a floor by applying a horizontal force of 8.0 N. The kinetic frictional force acting on the box has magnitude 5.0 N. As the box is dragged through 0.70 m along the way, what are (a) the work done by the collie’s applied force and (b) the increase in thermal energy of the bed and floor? >[!Example]- Module 8-4 Problem 45 >A rope is used to pull a 3.57 kg block at constant speed 4.06 m along a horizontal floor. The force on the block from the rope is 7.68 N and directed $15^\circ$ above the horizontal. What are (a) the work done by the rope’s force, (b) the increase in thermal energy of the block – floor system, and (c) the coefficient of kinetic friction between the block and floor? >[!Example]- Module 8-5 Problem 58 >A cookie jar is moving up a $40^\circ$ incline. At a point 55 cm from the bottom of the incline (measured along the incline), the jar has a speed of 1.4 m/s. The coefficient of kinetic friction between jar and incline is 0.15. (a) How much farther up the incline will the jar move? (b) How fast will it be going when it has slid back to the bottom of the incline? (c) Do the answers to (a) and (b) increase, decrease, or remain the same if we decrease the coefficient of kinetic friction (but do not change the given speed or location)? >[!Example]- Module 8-5 Problem 62 >n Fig. 8-55, a block slides along a path that is without friction until the block reaches the section of length $L = 0.75 m$, which begins at height $h = 2.0 m$ on a ramp of angle $u = 30^\circ$. In that section, the coefficient of kinetic friction is 0.40. The block passes through point A with a speed of 8.0 m/s. If the block can reach point B (where the friction ends), what is its speed there, and if it cannot, what is its greatest height above A? # What's next? You're done! Hope you aren't feeling too frazzled yet. To navigate to our next instalment, Friction and Drag: [[Friction and Drag (Mechanics)]] To navigate to other physics topics: [[The (Incomplete) Physics Almanac]] To navigate to other mechanics' topics: [[Mechanics - A Contents Page]] To visit the thermodynamics pages for a continuation of potential energy, go here: [[Thermodynamics - A Contents Page]]