aleksxxx
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Okay -
This isn't really a homework problem, but i had a question about the work equation, W=Fdcos(theta)
Say, for example, you have a block sitting on a surface and there is friction, but we are not talking it into account.
There is a force of 10N applied parallel to the displacement of the box, which is 10m
I thought the work done would be 100N, becuase the Cos between displacement and force is 1.
My Kaplan instructor was saying that in that situation the work done would be zero, because you are using F=mg and the discplacement of the box and the cos between those are 90 deg. ==> W=mg(d)cos(90)=0
I know its really elementary, but i just could have sworn that F in the equation was force APPLIED.
thanks.
This isn't really a homework problem, but i had a question about the work equation, W=Fdcos(theta)
Say, for example, you have a block sitting on a surface and there is friction, but we are not talking it into account.
There is a force of 10N applied parallel to the displacement of the box, which is 10m
I thought the work done would be 100N, becuase the Cos between displacement and force is 1.
My Kaplan instructor was saying that in that situation the work done would be zero, because you are using F=mg and the discplacement of the box and the cos between those are 90 deg. ==> W=mg(d)cos(90)=0
I know its really elementary, but i just could have sworn that F in the equation was force APPLIED.
thanks.