Work due to Friction equals Change in Mechanical Energy Problem by Billy

[@chaosend3815]

At 4:15 how did you get tan(theta)? How do you go from cos(theta) x (hf/sin(theta)) to sin(theta)/cos(theta) x (hf) ?

[@FlippingPhysics]

I think you might have missed that tan(theta) is in the denominator. Does that help?

[@chaosend3815]

Not very much. I don't understand how the cos(theta) got below sin(theta) to form tan(theta). I see it this way: cosx (hf/sinx) --> (cosx/sinx) (hf) --> cotx (hf) --> (1/tanx) (hf) Basically simplifying to cotx and then change it to its reciprocal, (1/tanx). But I don't think this is the way you did it.

[@FlippingPhysics]

(cosθ·hf)/sinθ = hf/tanθ

[@kainwalker7287]

1 divided by 3/2 can be written as 2/3 (2 of 3/2 goes up). In the same way, 2/3 can be written as 1 divided by 3/2. This is what he did in the video. Cos(theta)/sin(theta) can be written as 1 divided by sin(theta)/cos(theta). Sin(theta)/cos(theta) equals to tan(theta). So, It can be written as 1/tan(theta). Hope It clears your doubt^^

[@SaraPeckham]

Is Cos theta over Sine theta at 4:25 Tan theta?

[@FlippingPhysics]

Not quite. Tangent of theta equals sine theta divided by cosine theta. However, in this problem we have cosine theta divided by sine theta which equals 1 divided by tangent theta. Hope that helps!

[@andrewjustin256]

In that gigantic quantity of 1+ u/tan ©, I also had g with tangent. I solved the problem by breaking it down into two parts, elastic potential energy to kinetic before reaching the incline, and then the kinetic energy before the incline to potential energy on the incline, evidently with the friction and mechanical energy equation. But I have that extra g.

[@jennaalqdah9538]

These videos are amazing. Thank you so much

[@FlippingPhysics]

You are welcome!

[@saavankaneria1497]

When calculating work ( W=Fdcos(theta)) why is the cos(theta) necessary?

[@FlippingPhysics]

Because that is how work is defined in physics. The cosine of theta makes it so only the force in the direction of the displacement is included in the work equation. I would suggest you watch my introductory work video which walks through examples with different angles. www.flippingphysics.com/work-intro.html

[@robindixon2949]

I assigned this video to my students today. Let's see if they like this comment XD

[@FlippingPhysics]

Good luck!

[@Stravilight]

really only 4 comments? This vid is good!

[@FlippingPhysics]

Thanks. I'll pass your kind words on to Billy.

[@Apwolsopcjrhei]

Cool, this really filled my brain, thanks!

[@FlippingPhysics]

+Brianmax8HC I was hoping to expand the brain's capacity rather than filling it. Sorry 'bout that.

[@Apwolsopcjrhei]

+Flipping Physics Yeah, I'm not good at English. Thanks for teaching me some english phrases

[@FlippingPhysics]

+Brianmax8HC Your English is fine. It would be very typical to say that your "brain is full". I just decided to turn the phrase around and bit and wish for an expanded brain rather than a full one.

[@deepthrought]

That was really helpful thanks a lot

[@FlippingPhysics]

You are welcome. I will pass the word on to Billy.

[@BanValsimot]

Amazing videos

[@FlippingPhysics]

Thanks!