Water skiers often ride to one side of the center line of a boat, as shown in the figure . In this case, the ski boat is traveling at 15 {\rm m}/{\rm s} and the tension in the rope is 76 N.
If the boat does 3600 J of work on the skier in 53.6 m, what is the angle \theta between the tow rope and the center line of the boat?
To find the angle θ between the tow rope and the center line of the boat, we can use the equation for work done:
Work = Force * Distance * cos(θ)
In this case, the work done by the boat on the skier is given as 3600 J, the tension in the rope is 76 N, and the distance covered is 53.6 m. We can rearrange the equation to solve for θ:
θ = cos^(-1)((Work) / (Force * Distance))
Substituting the given values:
θ = cos^(-1)((3600 J) / (76 N * 53.6 m))
Now we can calculate the angle θ using a scientific calculator:
θ ≈ cos^(-1)(0.850)
Using a calculator, we find that the inverse cosine of 0.850 is approximately 31.8 degrees.
Therefore, the angle θ between the tow rope and the center line of the boat is approximately 31.8 degrees.
To find the angle θ between the tow rope and the center line of the boat, we can use the work-energy principle:
Work done on an object = Change in kinetic energy
The work done on the skier is given as 3600 J, and we need to find the angle θ. We know that the boat is traveling at a speed of 15 m/s and the tension in the rope is 76 N.
The work done on the skier is given by the formula:
Work = Force x Distance x cos(θ)
where force is the tension in the rope and distance is the distance traveled by the skier.
Plugging in the given values, we have:
3600 J = 76 N x 53.6 m x cos(θ)
Rearranging the equation, we get:
cos(θ) = 3600 J / (76 N x 53.6 m)
cos(θ) ≈ 0.905
To find the angle θ, we take the inverse cosine (or arccos) of both sides:
θ = arccos(0.905)
Using a calculator, we find:
θ ≈ 24.5 degrees
Therefore, the angle θ between the tow rope and the center line of the boat is approximately 24.5 degrees.