When I look at W=Fcos(theta)d I want to use the F = 60N from the question. Should I assume that there should be an f subscript for the F in the equation, but was not written? Since Ff = (mew)*Fn.
Great video! I'm just confused about one thing. For the last example question, when you calculated work performed by the pulling force why did you use 60N and not the x component of the force which is 60cos30 (as displacement is in the x direction)? Really struggling with that point.
Hello Amar! If you go back and look the last example was a two-part question. The 1st part asked for the work performed by the pulling force in which case I did indeed use 60cos30. But the 2nd part asked for the work performed by friction. As friction opposed the motion it was in the exact opposite direction of the displacement which is why cos180 was used. Hope this helps!
Thank you for speaking at a normal speed. All of the school lectures have to be sped up 1.5-2x. It's so easy to follow along this way. Thank you!
You're welcome.
Chad is saving my grades
Glad to hear it!
When I look at W=Fcos(theta)d I want to use the F = 60N from the question. Should I assume that there should be an f subscript for the F in the equation, but was not written? Since Ff = (mew)*Fn.
Great video! I'm just confused about one thing. For the last example question, when you calculated work performed by the pulling force why did you use 60N and not the x component of the force which is 60cos30 (as displacement is in the x direction)? Really struggling with that point.
Hello Amar! If you go back and look the last example was a two-part question. The 1st part asked for the work performed by the pulling force in which case I did indeed use 60cos30.
But the 2nd part asked for the work performed by friction. As friction opposed the motion it was in the exact opposite direction of the displacement which is why cos180 was used.
Hope this helps!
@@ChadsPrep Ah okay! This makes so much more sense. Thank you!