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

    and 33175 was the answer

    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

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