college physics helpasap
posted by JU .
The water skier in the figure is at an angle of 35degree with respect to the center line of the boat, and is being pulled at a constant speed of 15m/s .
1)If the tension in the tow rope is 90.0N, how much work does the rope do on the skier in 30.0s ?
2)How much work does the resistive force of water do on the skier in the same time?
1) i did p=F.V 15*90=1260 then w=pt so 1260*30s and the answer to that was wrong
so lost i need help please

The force component along the direction of motion is T cos 35 = 63.0 N. Only the force component in that direction can do wrk. In 30 seconds the water skier moves 450 m. The Work done is 63 x 450 = 28,360 J

2) Since the speed does not change, all of the work done by the tow rope is used up overcoming friction. Friction does negative work of 28,360 J.

how did you get 450m? because the distance wasn't given in this problem

450 m is how far the water skier is pulled in 30 seconds at 15 m/s.
You chose to multiply power by time, which is OK, but you got the power wrong. 
ok thank you i got it. still baffled on how the distace is 450
but i multiplyed
90*450*cos35
w=f*d*costheta
and 33175 was the answer
thank you 
1)The tension of the rope in the x direction is 90*cos35 = 73.7
v=15m/s
t = 30
The equation for Power is W/t and F*v
Thus W/t=F*v
W=t*F*v
W=30*73.7*15
W=32850
2)the water does the same work in the opposite direction.
W=32850