Dynamics: Lesson 23 - Work and Energy Example Problem
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- Опубліковано 27 лют 2020
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I really hope you continue this series! Impulse and momentum would be great, I have a really bad dynamics professor and this whole COVID 19 stuff has made him even more useless. Your videos are a lifesaver and you've helped me so much with assignments and understanding! Thanks so much for all that you do, sending good vibes your way man
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thank goodness Dr. Hanson!!!! You uploaded this on perfect time, getting close to midterm for my dynamics class
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Perfect timing Dr. the exam is in 3 days. I’ve watched your videos since your static’s series and it is always extremely helpful
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Hey doctor Hanson the .5 should be squared right? Then the -90J for the Uspring makes sense. Also thank you so much for this video and all your videos for that matter, you are amazing and I tell people about you all the tine as you are the professor I dream about having for this class!
You are absolutely correct!
DR. Jeff Hanson, thank you for a classic selection and analysis of another solid example of Work and Energy in old fashion Dynamics. I really like these problems in Engineering Physics One and Dynamics.
I have come this far in my Dynamics course because of you. Please keep on putting out more content.
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Please please make a lecture on Impulse and Momentum
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This made a lot more sense then what my professor did in class. Thank you
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What about the friction force, I suppose we're assuming a frictionless plane? It would be a negative force acting against the path of the applied force, but I suppose you would need a coefficient of kinetic friction to solve and substitute into the "Fun" equation.
smooth incline is given
@@ruandre5619 I hear you, but 'smooth' doesn't necessarily equate to completely frictionless, which is obviously impossible.
@@paulatreides4574 I agree but I've seen that when they use that in textbooks, they imply a frictionless surface
exam at 5pm, cramming rn lets gooo
great video sir....big fan
also is 400n the net resultant force after all forces have been added? i thought for a second the sum of the forces which are Work=Fx -Wx-Fs where Fx(Force of the x component of 400N) Wx(gravity x component) and Fs(Force of the spring pulling back)
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On the work of the spring, he missed the ^2 for the equation ...-1/2(30)(0.5)^2.
sir why didn't u use the formula T1+EU1-2=T2?
Mr Hanson I want to begin by thanking you, you got me through statics. However, in lesson 22, the video before this one, you say the work of the force of friction on the crate is positive (even though the force vector goes against the direction of displacement), but in this lesson you say that the force acting against the crate (exactly same situation as last lesson) is negative....oh why god whyyyy?!?! i can not get a clear definition of when work is positive or negative and it is driving me nuts.
Work is defined as the scalar product of two vectors which means the infinitesimal distance, dx, or whichever plane you’re moving in, has a direction as well. If the angle between your force and motion is obtuse, then you’ll have negative work. This negative work means this force is working against the work you’re trying to achieve in a certain direction.
I think what the professor failed to say is that during pre-stretching, the spring force is pointing in the direction of displacement so it will do positive work. However, after it has been stretched at its final state, it is pointing in the opposite direction. If you distribute the negative sign, you’ll see that when it has been fully stretched, it will do negative work.
I think your explanation of the sign of work conflicts with your previous video.
in "freedom units".... 😂
actually the result is correct but , 0.5 must be squared to realize the answer
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welcome and enjoy ..
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Listening skill ..
i used nt components hhaha