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
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
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.
how did you get 450m? because the distance wasn't given in this problem
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.
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
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
To solve the first question, you correctly used the formula for power (P = F * V) to find the power exerted by the rope on the skier. However, to calculate the work done, you need to multiply the power by the time (W = P * t), not the power by itself again.
Let's calculate it step by step:
1) Find the power exerted by the rope:
P = F * V
P = 90.0N * 15m/s
P = 1350.0 W
2) Calculate the work done by the rope on the skier in 30.0 seconds:
W = P * t
W = 1350.0 W * 30.0s
W = 40,500.0 J (Joules)
So, the work done by the rope on the skier in 30.0 seconds is 40,500.0 Joules.
Now let's move on to the second question, which asks for the work done by the resistive force of water on the skier in the same time.
To calculate this, we need more information. Is there any additional data provided regarding the resistive force of water or any other relevant information about the setup?