Question

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

• A. 80 N
• B. 60 N
• C. 50 N
• D. 40 N

Answer 1

The correct option is C 50 N

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.

$2=\frac{(60-b)}{(40-b)}$

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:

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