# college physics helpasap

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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

• college physics helpasap - ,

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

• college physics helpasap - ,

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.

• college physics helpasap - ,

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

• college physics helpasap - ,

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.

• college physics helpasap - ,

ok thank you i got it. still baffled on how the distace is 450
but i multiplyed
90*450*cos35
w=f*d*costheta

thank you

• college physics helpasap - ,

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