Question

An ideal spring, with a pointer attached to its end, hangs next to a scale. With a 100-N weight attached, the pointer indicates “40” on the scale as shown. Using a 200-N weight instead results in “60” on the scale. Using an unknown weight X instead results in “30” on the scale. The weight of X is

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Answer 1

Answer:

By the Hooke's law says the force of a spring is a linear relationship, F=kx

The spring will have some reading, b, when it is unstretched. We need the raw distance the spring stretches, x. So it follows that the raw distance x = the reading (the pointer) - b.

So, using x=reading−b

So, using x = reading - b (e.g., 40-b is the first distance the spring is stretched), we write the following system of equations:

100N=k(40−b)

200N=k(60−b)

We can divide them by each other if we're so inclined.

2=

(40−b)

(60−b)

Solving for b, we can get the unstretched length of the spring. b = 20.

Now, we need to find k.

100N=k(40−20)

k=5.

So, what is the force that corresponds to a reading of 30?

Well:

F=kx=5(30−b)=5(30−20)=50N.

Answer 2

Given:

With a 100-N weight attached, the pointer indicates “40” on the scale as shown. Using a 200-N weight instead results in “60” on the scale.

To find:

The unknown weight X which indicates “30” on the scale.

Solution:

"Hooke's law states that the strain of the material is proportional to the applied stress within the elastic limit of that material ( F=kx )."

We know that raw distance = the reading (the pointer - b).

According to the question,

100N=k(40−b)

 200N=k(60−b)

On solving the above equations, we get k = 5, b = 20

F = kx = 5(30 - 20) = 50 N.

Therefore, the weight of X is 50 N.