- #1
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!
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
W=Fd=F(cos[tex]\vartheta[/tex])d
v=d/t
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[tex]\vartheta[/tex])d
W=(100N)(cos25)(180m)
=16.3kJ
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[tex]\vartheta[/tex])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[tex]\vartheta[/tex])d
W=(100N)(cos25)(180m)
=16.3kJ