Solve Work & Angle Homework: Plane Towing Glider & Water Skier

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SUMMARY

The discussion focuses on solving two physics problems involving work and angles related to towing a glider and water skiing. For the plane towing the glider, 2.00 x 105 J of work is done over 145 m with a tension of 2560 N, resulting in an angle of 57.4 degrees between the tow rope and the horizontal. In the second problem, a ski boat exerts 3500 J of work on a skier over 50 m with a rope tension of 75 N, yielding an angle of 21.04 degrees between the tow rope and the center line of the boat. The speed of the boat is deemed irrelevant for both problems since the calculations are based solely on work.

PREREQUISITES
  • Understanding of work-energy principles in physics
  • Familiarity with the formula W = F d cos θ
  • Knowledge of tension forces in towing scenarios
  • Basic trigonometry for angle calculations
NEXT STEPS
  • Study the concept of power in relation to work and speed
  • Explore the implications of angle in towing mechanics
  • Learn about the relationship between work, energy, and motion in physics
  • Investigate real-world applications of towing forces in aviation and water sports
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts of work and angle in practical scenarios.

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Homework Statement


1. A small plane tows a glider at constant speed and altitude. If the plane does 2.00 x 105 J of work to tow the glider 145 m and the tension in the tow rope is 2560 N, what is the angle between the tow rope and the horizontal?

2. Water skiers often ride to one side of the center line of a boat as shown below. In this case, the ski boat is traveling at 15 m/s and the tension in the rope is 75 N. If the boat does 3500 J of work on the skier in 50.0 m, what is the angle θ between the tow rope and the center line of the boat?
untitled-10.jpg


Homework Equations


W = F d cos θ
W = Δ KE


The Attempt at a Solution


1.
W = F d cos θ
2 x 105 = 2560 x 145 cos θ
θ = 57.4o
I'm not sure because I don't know the picture of plane tows a glider. I am only using the formula

2.
W = F d cos θ
3500 = 75 x 50 cos θ
θ = 21.04o
Is the speed given in the question not important?

Thanks
 
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All looks good to me. The speed is irrelevant in part b.
 
and me. Speed irrelevant in both parts as the problem is the same.

Speed information would have be needed had the question given you "power" rather than "work".
 
TSny said:
All looks good to me. The speed is irrelevant in part b.

CWatters said:
and me. Speed irrelevant in both parts as the problem is the same.

Speed information would have be needed had the question given you "power" rather than "work".

Thanks :smile:
 

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