Tension: Force + Velocity. Breaking Strength.

AI Thread Summary
The discussion focuses on calculating the minimum breaking strength of a fishing line needed to stop a 90 N salmon moving at 2.5 m/s over a distance of 18 cm. Initially, the user underestimated the breaking strength, thinking it only needed to exceed 90 N. However, they realized that the fish's horizontal movement and the need to decelerate to a stop would increase the required force. By applying kinematics to find acceleration and using the weight of the fish to derive its mass, they calculated that the necessary breaking strength is approximately 159.44 N. This solution highlights the importance of considering both force and velocity in tension-related problems.
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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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