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Nadya [2.5K]
4 years ago
5

A fisherman is fishing from a bridge and is using a "44.0-N test line." In other words, the line will sustain a maximum force of

44.0 N without breaking. What is the weight of the heaviest fish that can be pulled up vertically, when the line is reeled in (a) at constant speed and (b) with an acceleration whose magnitude is 1.26 m/s2?
Physics
1 answer:
avanturin [10]4 years ago
7 0

Answer:

(a) W= 44N

(b)W= 31.65 N

Explanation:

Data

T=44 N : Maximum force that the rope can withstand without breaking

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

(a)  We apply the formula (1) at constant speed , then, a=0

W: heaviest fish that can be pulled up vertically

∑F = 0

T-W =0  

W = T

W= 44N

(b)  We apply the formula (1) , a= 1.26 m/s²

W: heaviest fish that can be pulled up vertically

W= m*g

m= W/g

g= 9.8 m/s² : acceleration due to gravity

∑F = 0

T-W = m*a

T= W+(W/g)*a

44=W*(1+1/9.8)* (1.26 )

44= W* 1.39

W= 44/1.39

W= 31.65 N

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Explanation:

Conceptual analysis

We apply the free fall formula for position (y) and speed (v) at any time (t).

As gravity opposes movement the sign in the equations is negative.:  

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y: The vertical distance the ball moves at time t  

vi: Initial speed

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Known information

We know the following data:

Vi=15 m / s

g =9.8 \frac{m}{s^{2} }

t=1s ,1.5s,2s

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We replace t in the equations (1) and (2)  

a) t=1s

y = 15*1 - ½ 9.8*1^{2}=15-4.9=10.1m

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b) t=1.5s

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c) t=2s

y = 15*2 - ½ 9.8*2^{2}= 30-19.6=10.4 m

v=15-9.8*2 =15-19.6=-4.6m/s  (The minus sign (-) indicates that the ball is already going down)

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