The graph shows the net external force component Fcos(theta) along the displacement as a function of the magnitude of the displacement s. The graph applies to a 56 kg ice skater.
link to graph: http://i48.photobucket.com/albums/f226/buttrefly18/06_71.gif
(a) How much work does the net force component do on the skater from 0 to 3.0 m?
(b) How much work does the net force component do on the skater from 3.0 m to 6.0 m?
(c) If the initial speed of the skater is 2.5 m/s when s = 0, what is the speed when s = 6.0 m?
2007-03-28 11:53:52 · 1 answers · asked by softball579 1 in Science & Mathematics ➔ Physics
Ok, I watched the graphic, and the graphic :
Remember : Work = Fcos(theta)*distance.
In the case of your graphic, it will be the area down the curve :
a) W = 31*3 = 96 Joules
b) from 3.0 to 6.0; W = cero = no area
c) Let's use the energy - work theorem :
initial energy - final energy = work done by the force
56*2.5^2 / 2 - 56*v^2 / 2 = 96
79 = 56*v^2 / 2
v = 1.67 m / s
2007-03-28 12:00:53 · answer #1 · answered by anakin_louix 6 · 0⤊ 0⤋