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+ | ~~SLIDESHOW~~ | ||

+ | |||

+ | ====== Chapter 9 - Energy in Nonisolated Systems ====== | ||

+ | |||

+ | ===== Work ===== | ||

+ | |||

+ | Work is a measure of what a force achieves, so when we calculate the work done by a force on an object we consider only it's displacement in the direction of the force. | ||

+ | |||

+ | Mathematically, as we consider both force $\vec{F}$ and displacement $\vec{d}$ to be vectors, and the work done on an object to be a scalar $W$, we say that the work is the scalar product, or dot product of force and displacement. We should note that this is not the displacement from an arbitrary origin, but rather the change in an objects position during a motion. | ||

+ | |||

+ | For a constant force | ||

+ | |||

+ | $W= \vec{F} \cdot \vec{d} $ | ||

+ | |||

+ | If the two vectors are at an angle $\theta$ to each other then we can say | ||

+ | |||

+ | $W = Fd\cos\theta$ | ||

+ | |||

+ | {{fdcostheta.png}} | ||

+ | |||

+ | ===== Work on a backpack ===== | ||

+ | |||

+ | Consider a hiker that walks first on the flat and then up a hill carrying a backpack, with a **constant velocity** on each section. | ||

+ | |||

+ | {{fonbp.png}} | ||

+ | |||

+ | A hiker carries a 10kg backpack 500 m on the flat at a constant velocity. How much work do they do on the backpack? | ||

+ | |||

+ | The hiker now climbs up a mountain that has a 5 degree incline. When he has walked 500m and still moving at the same speed how much work has he done on the backpack? | ||

+ | |||

+ | ===== 9.P.004 ===== | ||

+ | |||

+ | ===== More on the dot product ===== | ||

+ | |||

+ | The scalar product is commutative (order is not important) | ||

+ | |||

+ | $\vec{A}\cdot\vec{B}=\vec{B}\cdot\vec{A}$ | ||

+ | |||

+ | and distributive | ||

+ | |||

+ | $\vec{A}\cdot(\vec{B}+\vec{C})=\vec{A}\cdot\vec{B}+\vec{A}\cdot\vec{C}$ | ||

+ | |||

+ | The scalar product of two vectors written in unit vector notation is straightforward to calculate because the different unit vectors are perpendicular to each other. | ||

+ | |||

+ | $\hat{i}\cdot\hat{i}=\hat{j}\cdot\hat{j}=\hat{k}\cdot\hat{k}=1$ | ||

+ | |||

+ | $\hat{i}\cdot\hat{j}=\hat{i}\cdot\hat{k}=\hat{j}\cdot\hat{k}=0$ | ||

+ | |||

+ | So for two vectors | ||

+ | |||

+ | $\vec{A}=A_{x}\hat{i}+A_{y}\hat{j}+A_{z}\hat{k}$\\ | ||

+ | $\vec{B}=B_{x}\hat{i}+B_{y}\hat{j}+B_{z}\hat{k}$ | ||

+ | |||

+ | $\vec{A}\cdot\vec{B}=A_{x}B_{x}+A_{y}B_{y}+A_{z}B_{z}$ | ||

+ | |||

+ | ===== 9.P.011 ===== | ||

+ | |||

+ | ===== 9.P.016 ===== | ||

+ | |||

+ | |||

+ | |||

+ | ===== A more general definition of work ===== | ||

+ | |||

+ | Our previous definition of work was rather limited, in that it considered only a constant force. To be able to consider the work done by a force which can vary over a given path between two points in space we should better write | ||

+ | |||

+ | $W=\int_{r_{i}}^{r_{f}}\vec{F}\cdot d\vec{r}$ | ||

+ | |||

+ | Here we broke our path up into infintessimcal elements $d\vec{l}$ over which we can then integrate | ||

+ | |||

+ | as | ||

+ | |||

+ | $\vec{F}=F_{x}\hat{i}+F_{y}\hat{j}+F_{z}\hat{k}$\\ | ||

+ | $d\vec{r}=dx\,\hat{i}+dy\,\hat{j}+dz\,\hat{k}$ | ||

+ | |||

+ | This vector integral can be easily split in to separate integrals for the different components | ||

+ | |||

+ | $W=\int_{x_{i}}^{x_{f}}F_{x}\,dx+\int_{y_{i}}^{y_{f}}F_{y}\,dy+\int_{z_{i}}^{z_{f}}F_{z}\,dz$ | ||

+ | |||

+ | ===== 9.P.026 ===== | ||

+ | |||

+ | |||

+ | ===== Conservative Forces ==== | ||

+ | |||

+ | A force can be considered to be conservative, if the work done on the object by the force depends only on the beginning and end point of the motion and is independent of the path taken. | ||

+ | |||

+ | ===== Gravity is a conservative force ===== | ||

+ | |||

+ | {{conservativegravity.png}} | ||

+ | |||

+ | $W_{G}=\int_{1}^{2} \vec{F}_{G}\cdot d\vec{r}$ | ||

+ | |||

+ | Now as $\vec{F}_{G}=-mg\hat{j}$ and $d\vec{r}=dx\hat{i}+dy\hat{j}+dz\hat{k}$ | ||

+ | |||

+ | $\vec{F}_{G}\cdot d\vec{l}=-mg\,dy$ | ||

+ | |||

+ | and | ||

+ | |||

+ | $W_{G}=-\int_{y_{1}}^{y_{2}}mg\,dy=-mg(y_{2}-y_{1})$ | ||

+ | |||

+ | The work done depends only the change in height. | ||

+ | |||

+ | ===== Friction is a non-conservative force ===== | ||

+ | |||

+ | A force can only be conservative if the net work done by the force on an object moving around any closed path is zero. | ||

+ | |||

+ | This means that for an object moving from point 1 to point 2 under force the work must be equal and opposite for when it moves from point 2 to point 1. | ||

+ | |||

+ | As we know, the friction force is always in the opposite direction to motion, if we push an object a distance against a frictional force $\vec{F_{Fr}}$ then the work done by the friction on the object is $-F_{Fr}d$. (Note that the work done is negative, the frictional force reduces the kinetic energy.) If we push it back to it's original position the force is now in the opposite direction so the work done by the frictional force is again $-F_{Fr}d$. As the total work done is $-2F_{Fr}d$ we can see that friction is not a conservative force. | ||

+ | |||

+ | ===== Work-Mechanical Energy theorem ===== | ||

+ | |||

+ | The work-mechanical energy theorem says that the net work done on an system $W_{tot}$ is equal to the change in the total mechanical energy $\Delta E$ | ||

+ | |||

+ | $W_{tot}=\Delta E$ | ||

+ | |||

+ | $E=K+U$ | ||

+ | |||

+ | $W_{tot}=\Delta K+ \Delta U$ | ||

+ | |||

+ | |||

+ | |||

+ | |||

+ | |||

+ | ===== Dropping a weight on a nail demo ===== | ||

+ | |||

+ | | {{naildrop.png?100}} | During the drop of the weight only conservative forces act. So whatever potential energy we lose, $mg\Delta h$, gets converted in to kinetic energy, $\frac{1}{2}mv^{2}$. \\ \\ If we drop the weight from twice the height we see that the ratio of the kinetic energy should be $\frac{K_{2h}}{K_{h}}=2$ and as the kinetic energy is proportional to the square of the velocity the ratio $\frac{v_{2h}^{2}}{v_{h}^2}=2$ and $\frac{v_{2h}}{v_{h}}=\sqrt{2}$ \\ \\ When the weight hits the nail the force is used to push the nail in to the block. We should note that this a non-conservative force, the force is used to deform the block in an irreversible fashion! However the work-energy theorem still applies, so the amount of work done is equal to the kinetic energy lost. \\ \\ As $W=Fd$ the distance the nail goes in should be twice as much when the kinetic energy is twice as much, $\frac{d_{2h}}{d_{h}}=2$. | | ||

+ | |||

+ | |||

+ | ===== Internal Energy===== | ||

+ | |||

+ | Besides kinetic energy and potential energy, a system composed of real objects possesses various forms of internal energy. One obvious form of energy is the thermal energy associated with the temperature of an object. There is also energy associated with the chemical bonds in the material. | ||

+ | |||

+ | When we drove the nail in to the wood we both increased the temperature of the nail and the wood and broke many bonds in the wood, and this is where most of the energy went. | ||

+ | |||

+ | If we were able to quantify this internal energy we could make a more general statement about work and energy, which we call the work energy theorem | ||

+ | |||

+ | $E_{i}+W_{tot}=E_{f}+\Delta E_{int}$ | ||

+ | |||

+ | or | ||

+ | |||

+ | $K_{i}+U_{i}+W_{tot}=K_{f}+U_{f}+\Delta E_{int}$ | ||

+ | |||

+ | ===== 9.P.041 ===== | ||

+ | |||

+ | |||

+ | |||

+ | ===== Snow Bike ===== | ||

+ | |||

+ | [[http://www.youtube.com/watch?v=Ub-FSE8FcyA&feature=related|Snow Bike Video]] | ||

+ | |||

+ | {{snowbike.png}} | ||

+ | |||

+ | The work-energy theorem can be used for this problem. When a frictional force $\vec{F}_{fr}$ is acting and an object travels a distance $d$ while it goes from a height $h_{1}$ to $h_{2}$, changing it's velocity from $v_{1}$ to $v_{2}$, the work-energy theorem tells us | ||

+ | |||

+ | $\frac{1}{2}mv_{1}^2+mgh_{1}=\frac{1}{2}mv_{2}^2+mgh_{2}+F_{fr}d$ | ||

+ | |||

+ | If the cyclist starts from rest then what is his velocity at the bottom of the hill? | ||

+ | |||

+ | How far will he travel down the road before coming to a stop? | ||

+ | |||

+ | ===== Power ===== | ||

+ | |||

+ | Power is defined as the rate at which work is done. | ||

+ | |||

+ | Average power for some work $W$ performed in a time interval $\Delta t$ is given by | ||

+ | |||

+ | $P_{av}=\frac{W}{\Delta t}$ | ||

+ | |||

+ | Instantaneous power is given by | ||

+ | |||

+ | $P=\frac{dW}{dt}$ | ||

+ | |||

+ | Recall that $dW=\vec{F}\cdot d\vec{l}$, which means that | ||

+ | |||

+ | $P=\frac{dW}{dt}=\vec{F}\cdot \frac{d\vec{l}}{dt}=\vec{F}\cdot\vec{v}$ | ||

+ | |||

+ | Units of Power are J/s or W (Watts). | ||

+ | |||

+ | ===== 9.P.065 ===== | ||