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If the same fish is attached to the end of the unstretched spring and then allowed to fall from rest, through what maximum distance does it stretch the spring? (Hint: Calculate the force constant of the spring in terms of the distance d and the mass m of the fish.)

User Jullie
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1 Answer

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Answer:

x=2d

Step-by-step explanation:

initial stretch in the spring is d

so using Hook's law

at equilibrium position

k×d=mg

where k= spring constant

m= mass of fish

g= acceleration due to gravity.

d=mg/k ................ (1)

in second case by energy conservation

1/2 kx^2=mgx

x=2mg/k

using equation 1

x=2d

User Uzzar
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