Principle of Work and Energy Example 2- Engineering Dynamics

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"You could change your major and that would be negative work" You should do standup

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[@structurefree]

Thanks.

[@htebreeder]

Thanks for the video!

[@ChieefJeef]

how come you dont consider gravitational potential energy?

[@structurefree]

in the principle of work and energy, we're dealing with work. the work of weight is related to the gravitational potential energy.

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[@mattharper3972]

wouldn't it be more efficient to use a potential energy term in your initial equation and set state 2 as a y=0 datum? then you'd have no PE in state 2 but you'd have PE in state 1 instead of a work done by gravity term. I know it's 6 of one half a dozen of the other just my thoughts. Really enjoyed the video though. Good brush up on skills before my final exam.

[@structurefree]

+Matt Harper thanks for the suggestion and comment!

[@zee9000_]

You should do video going over the theory - examples are limited. Good job anyway

[@libsingburg]

Is your total work supposed to be +0.30mgcos(15)(d) [since your friction is in the positive sense of the x component] and -mg sin (15) (d) [ since your weight is in the negative sense of the x component] ? Please enlighten me. thanks!

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[@engrc9185]

Why didn't you include the force of inertia?

[@mattharper3972]

+Trich Candelario having Kinetic energy in both states is doing that. This is a conservation of energy problem. You don't need an inertial force term.

[@structurefree]

+Matt Harper thank you! I love it when others share their knowledge in the comments!!

[@ezalcinoryahoocom]

Why is it when you summed the the forces in Y you got cos(15). When I did it I got sin(15). What sort of witch craft happened there, may I ask.

[@TheLittlepea21]

I think he made an error because in my summation of the forces i got mgsintheta

[@ezalcinoryahoocom]

TheLittlepea21 yeah, ok. maybe its an error.

[@structurefree]

Elyze IsAwsome Thank you for your interest and question. When I sum forces in the Y, I get Np - W cos(15) = 0. If you break up the W force vector into a vector triangle with Wx and Wy components (where X is parallel to the incline and Y is perpendicular to it), the 15 deg angle is at the intersection of Wy and W (aka the hypotenuse) . That is why, Wy = W cos(15). Hope that helps.

[@ezalcinoryahoocom]

ahhhhh I see what your saying, i had a slow moment.

[@jvalpatel2076]

I doubt this will get answered but fuggit. When you split F=ma into 2 equations to try and solve for other variables. Why is the equation for Fy -> -Wcos15 + N=0 instead of -Wsin15 + N = 0 since we are dealing with the Y axis?

[@osmick15]

So why exactly Np associated with cos, I am a little lost? thanks

[@obarbier]

WCOS15=y component

[@osmick15]

olivier barbier I had a really bad brain fart. Thanks man.

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[@Bey0ndestiny]

Why do we assume that sum of forces at y direction is 0?

[@SuccessfulLoser]

+Yen Fung There's no acceleration in the y-direction. So a=0.

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what happends when any elastics force is acting on the body ?? like if the body is connected with spring ? kindly upload the video in which elastic or externl force is applied ... waiting

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