Starting at rest at the edge of a swimming pool, a 72.0-kg athlete swims along the surface of the water and reaches a speed of 1.20 m/s by doing the work Wnc1= +161 J. Find the nonconservative work, Wnc2, done by the water on the athlete.
Explanations would be appreciated. Thanks!
2007-03-27 08:02:26 · 3 answers · asked by 123haha 1 in Science & Mathematics ➔ Physics
They tell you how much work he did.
How much kinetic energy does he gain?
1/2 m v^2
The rest of the work is wasted (ie, nonconservative work done by water on swimmer).
2007-03-27 08:08:41 · answer #1 · answered by Anonymous · 0⤊ 0⤋
These contributors before me are on the right track.
Well since you've asked then here we go.
Work done by the athlete equal to his kinetic energy + resistance of water
Wnc1=Ke+Wnc2
Wnc2,=Wnc1-Ke
Wnc2 =Wnc1 - 0.5 mV^2
Wnc2=+161 - 0.5 x 75 x (1.2)^2=
Wnc2=107Joules
2007-03-27 08:56:23 · answer #2 · answered by Edward 7 · 1⤊ 0⤋
The work done by the water is the work done by the swimmer - the kinetic energy hes got. So its 161 - 0.5*72*(1.2^2) which is 109.16 joules.
2007-03-27 08:42:28 · answer #3 · answered by Anonymous · 1⤊ 0⤋