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8 days ago

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A scale used to weigh fish consists of a spring hung from a support. The spring's equilibrium length is 10.0 cm. When a 4.0 kg f

ish is suspended from the end of the spring, it stretches to a length of 13.4 cm.
(Part A) What is the spring constant k for this spring? Express your answer with the appropriate units.
(Part B) If an 8.0 kg fish is suspended from the spring, what will be the length of the spring? Express your answer with the appropriate units.

1 answer:

ValentinkaMS [2.4K]8 days ago

5 0

A) 1153 N/m

We determine the spring constant using Hooke's law:

$F=kx$

where

F represents the force exerted on the spring

k denotes the spring constant

x signifies the spring's displacement

In this case, a fish with mass m = 4.0 kg is hanging from the spring, and hence, the force applied is the fish's weight:

$F=mg=(4.0 kg)(9.8 m/s^2)=39.2 N$

and the spring's displacement is:

$x = 13.4 cm - 10.0 cm = 3.4 cm = 0.034 m$

thus the spring constant amounts to

$k=\frac{F}{x}=\frac{39.2 N}{0.034 m}=1153 N/m$

B) 16.8 cm

For the second situation, a fish of mass

m = 8.0 kg

is attached to the spring. Hence, the force applied to the spring is

$F=mg=(8.0 kg)(9.8 m/s^2)=78.4 N$

Thus, we can ascertain the spring's displacement:

$x=\frac{F}{k}=\frac{78.4 N}{1153 N/m}=0.068 m = 6.8 cm$

Considering the spring's equilibrium length of

$x_0 = 10.0 cm$

the new length of the spring becomes

$x' = 10.0 cm + 6.8 cm = 16.8 cm$

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