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phy131studiof16:lectures:chapter7 [2016/07/28 15:45] mdawber [Gravitational Potential Energy] |
phy131studiof16:lectures:chapter7 [2016/07/28 15:46] (current) mdawber |
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- | ===== Nail Drop Demo ===== | ||

- | | {{naildrop.png?100}} | If I drop the weight from twice the height what is the ratio of the velocity measured at the photogate to the velocity measured in the first drop?\\ \\ If I drop the weight from twice the height what is the ratio of the distance the nail goes in to the distance it went in on the first drop? | | ||

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- | ===== Nail Drop Explained ===== | ||

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- | 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, $\frac{1}{2}mv^{2}$. | ||

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- | 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 $\frac{v_{2h}}{v_{h}}=\sqrt{2}$ | ||

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- | 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. | ||

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- | As $W=Fd$ the distance the nail goes in should be twice as much when the kinetic energy is twice as much, $\frac{d_{2h}}{d_{h}}=2$. | ||

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- | ===== Total Energy Conservation ===== | ||

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- | Non conservative forces remove mechanical energy from the system, but it is not destroyed, it is simply converted to a different form of energy (frequently, but not always, heat). | ||

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- | The total energy conservation law can also be useful, for example 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 $v_{2}$, conservation of total energy tells us | ||

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

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- | ===== Snow Bike ===== | ||

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

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- | {{snowbike.png}} | ||

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

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