Extreme Sports: Waterboarding and Physics (Work/Tension)

[minimax]

Hi, here's an interesting question on EXTREME SPORTS! (tried to make the title interesting)
hoping someone will verify my work/procedure in solving the que. or put me in the right direction :D Thank you!

Homework Statement

A waterboarder is at an angle of 25 degrees with respect to the straight central path of a top speed motor boat. She's being pulled at a constant speed of 18m/s. If the tension of her tow rope is 100N
i) How much work does the tope do on the boarder in 10.0s
ii) How much work does the resistive force of the water have on her in the same amount of time

Homework Equations

W=Fd=F(cos$$\vartheta$$)d
v=d/t

The Attempt at a Solution

i) v=d/t therefore d=vt
d=(10.0s)(18m/s)=180m travelled
W=Fd
=(100N)(180m)
=18kJ

ii) d=180m/s
W=Fd=F(cos$$\vartheta$$)d
W=(100N)(cos25)(180m)
=16.3kJ

 

[minimax]

I've added a diagram, hope this helps

 

[Moetasim]

Hi there! I am happy to help verify your work and provide feedback on your procedure. Your approach seems to be on the right track, but there are a few things that can be improved upon in your solution.

Firstly, in part i) of your solution, you correctly calculated the distance travelled by the waterboarder in 10.0 seconds. However, when calculating the work done by the tow rope, you used the formula W=Fd, but you forgot to include the angle between the force and displacement, theta (θ). This angle is important because the tension force is not acting in the same direction as the displacement of the waterboarder. The correct formula to use would be W=Fdcosθ. In this case, the angle between the force and displacement is 25 degrees, so the calculation should be:

W=(100N)(180m)cos25
= 16.3 kJ

In part ii) of your solution, you used the correct formula W=Fdcosθ to calculate the work done by the resistive force of the water. However, you used the wrong distance in your calculation. The distance travelled by the waterboarder in 10.0 seconds is 180m, not 180m/s. So the correct calculation would be:

W=(100N)(180m)cos25
= 16.3 kJ

Overall, your approach and equations were correct, but be sure to double check your calculations and include all necessary variables in the formulas. Keep up the good work!