Posted by JU on .
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

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drwls,
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

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drwls,
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.

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

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drwls,
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. 
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JU,
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 
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Omer Syed,
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