(1/2)mv f 2 =(1/2)mv i 2 -f k d+W other forces suppose a 6kg block initially at
ID: 1696844 • Letter: #
Question
(1/2)mvf2=(1/2)mvi2-fkd+Wother forces
suppose a 6kg block initially at rest is pulled to the right along a horizontal surface by a constant horizontal force of 12N. find the speed of the block after it has moved 3.0m if the surfaces in contact have a coefficient of kinetic friction of .15.
so my question is what is "Wother forces" mean? is it the work of forces in just the x-> direction? or both the x, and y direction?
p.s. i know how to solve the question, im just wondering what "Wother forces "means......
many thanks!
Explanation / Answer
F d = W = 12 * 3 = 36 J work done in pulling block W = 1/2 m v^2 + Ff d where Ff is the work done by friction Ff d = u m g d or the energy lost to friction 1/2 m v^2 = F d - u m g d = 36 - .15 * 6 * 9.8 * 3 = 9.54 J 1/2 m v^2 = 9.54 v = (19.1 / 6)^1/2 = 1.78 m/s This is just work in = work out
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