Unformatted text preview: x = 0 – 15(cos50 o ) – 12(cos30 o ) = 20 F net y = 10 + 15(sin50 o ) – 12(sin30 o ) = 15.5 F net = (F net x 2 + F net y 2 )1/2 = 25.3N Ø = tan1 (15.5/20) = 142 o (2 nd quadrant) W = F net d cos Ø = (25.3) (0.4) cos (0 o )= 10.1J Work energy Theorem: ΔK = K f – K i = ∑WK f = m v i 2 /2 + net workW g = m g cos(0 o ) = m g...
View
Full Document
 Spring '09
 CampbellBrown
 Physics, Energy, Force, Kinetic Energy, Potential Energy, Work, vector component

Click to edit the document details