Tension: Force + Velocity. Breaking Strength.

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SUMMARY

The discussion focuses on calculating the minimum breaking strength of a fishing line required to stop a salmon weighing 90.0 N moving horizontally at an initial velocity of 2.5 m/s over a distance of 18.0 cm. The solution involves using kinematic equations to derive acceleration, resulting in an acceleration of 17.361 m/s². Subsequently, the mass of the fish is calculated to be 9.18 kg, leading to a required breaking strength of 159.443 N for the fishing line to effectively stop the fish.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with kinematic equations
  • Knowledge of force, mass, and acceleration relationships
  • Basic grasp of unit conversions (N to kg)
NEXT STEPS
  • Study kinematic equations in-depth, particularly the equation Vf² = Vi² + 2a(X - Xo)
  • Learn about the relationship between force, mass, and acceleration in physics
  • Explore real-world applications of breaking strength in materials science
  • Investigate the effects of velocity on force in dynamic systems
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in the mechanics of forces and motion, particularly in practical applications like fishing line strength calculations.

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[SOLVED] Tension: Force + Velocity. Breaking Strength.

Homework Statement


The tension at which a fishing line breaks is commonly called the ``breaking strength''. What minimum breaking strength is needed for a fishing line to stop a salmon that weighs 90.0 N in 18.0 cm if the fish is moving horizontally with an initial velocity of 2.5 m/s? Assume the acceleration is constant.

Homework Equations


F=mg, F=ma, V^2 = Vo^2 + 2a(X-Xo)

The Attempt at a Solution


I'm not really sure how to tackle this problem. At first I simply assumed the minimum breaking strength of the string would have to be at least > 90N. However the fact that it's moving at 2.5m/s says to me that the 18cm of distance traveled increases the force created by the fish. But I'm not sure, algebraically, how to relate Force to Velocity. Also, since the fish weighs 90N, that implies its weight in relation to gravity. How would this translate when analyzing only movement in the X direction?
 
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Ok. This problem is trivial and I apologize for wasting space on your forum, but maybe some day it will help someone else who gets caught up on the details...

I neglected to notice that final Velocity would be 0. Once I realized this I saw I could use one of the kinematics equations to derive acceleration.

Here's what I did...

(Vf^2-Vi^2)/(X-Xo) = 2a
--> 0m/s - (2.5^2m/s)/(0m - .18m) = 2a
--> a = 17.361m/s^2

Now I can derive the mass of the fish by using F in the y-direction (since we know gravity is 9.8m/s^2)...

F=mg
--> 90N = m(9.8m/s^2)
--> m = 9.18kg

Now, put it all together for the x-direction...

F = (9.18kg)(17.361m/s^2) = 159.443 N
 

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