How much work does the resistive force of water do on the skier in 10secs?

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The discussion centers on calculating the work done by the resistive force of water on a water skier being pulled at a constant speed. For part (a), the work done by the rope is calculated as 10,321.3 Joules using the formula W = (tension × cos(angle)) × (speed × time). In part (b), participants explore the relationship between the forces, noting that since the skier moves at a constant speed, the forces balance out, leading to F(rope) + F(water) = 0. There is confusion regarding the measurement of force in Joules, as work is typically expressed in Joules while force is in Newtons. The discussion emphasizes the need for clarity in understanding these physical concepts.
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i think i did part (a) right but for part (b) am clueless can someone help me out please?

Problem-

The water skier in Figure 7-14 is at an angle of 35 with respect to the center line of the boat, and is being pulled at a constant speed of 14m/s. If the tension in the ton rope is 90N, (a) how much work does the rope do on the skier in 10secs? (b) How much work does the resistive force of water do on the skier in the same time?

answer-

(a) W= (90N cos35) (14m/s)(10secs)
= 10321.3J

(b)
 
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Since the skier is moving at a constant speed, the sum of the forces in the system is equal to 0 so..

F(rope) + F(water) = 0
F(rope) = -F(water)

so.. how much force is applied over how much distance by the water?
 
so does the resistance force for water is it -10321.3J so that it is equal to zero?
 
Since when is force measured in Joules!?
 

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