Question

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 fish is suspended from the end of the spring, it stretches to a length of 12.0 cm.

What is the spring constant k for this spring?

If an 8.0 kg fish is suspended from the spring, what will be the length of the spring?
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Answer 1

We can use Hooke's Law to find the spring constant k:

F = kx

where F is the force applied to the spring, x is the displacement of the spring from its equilibrium length, and k is the spring constant.

We can find the force applied to the spring by using the weight of the 4.0 kg fish:

F = mg = (4.0 kg)(9.81 m/s^2) = 39.24 N

The displacement of the spring is the difference between its length with the fish and its equilibrium length:

x = 12.0 cm - 10.0 cm = 2.0 cm = 0.02 m

Now we can solve for k:

k = F/x = 39.24 N / 0.02 m = 1962 N/m

To find the length of the spring with an 8.0 kg fish suspended from it, we can use the same formula with the new weight:

F = mg = (8.0 kg)(9.81 m/s^2) = 78.48 N

We can solve for x, which is the new displacement of the spring:

x = F/k = 78.48 N / 1962 N/m = 0.04 m

Therefore, the length of the spring will be:

10.0 cm + 4.0 cm = 14.0 cm

Step-by-step explanation: