## Trending News

# I need help on a physics problem about work and springs. Help me please?

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:

### 4 Answers

- Steve TLv 41 decade agoFavorite Answer
Hooke's law says the force of a spring is a linear relationship, F = kx, if I remember correctly.

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

100 N = k(40 - b)

200 N = k(60 - b)

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

2 = (60-b)/(40-b)

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

Now, we need to find k.

100 N = 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) = 50 N.

30 - 20 = 10 is the distance the spring stretched. We multiply this by the spring constant k to get the force the spring is applying at that distance.

Hope that can help.

- Don MLv 71 decade ago
There are two formulas for force here that are in equlibrium: the weight pulling down and the spring pulling up.

For the weight:

F = the weight of the object in Newtons

For the spring (pulling up):

F = -kx (where k is the spring constant for this spring and x is the displacement. This MAY or MAY NOT be the number indicated on the scale.

Let a be the displacement for the first example and let b be the displacement for the second example. Because the spring constant is just that -- CONSTANT, the actual DISPLACEMENT (not the number on the scale, necessarily) under 200N should be twice that under 100N.

100N = -ka

200N = -kb

By this,

100/-a = 200/-b

100b = 200a

and b=2a

In the first example, the scale is marked 40 under displacement a.

In the second example, the scale is marked 60 under displacement b (which is 2a). We can figure out what the 'zero position' on the scale is from this.

Let the zero position be represented by z.

40 = z + a

60 = z + 2a

So 2(z+2a) = 3(z+a)

2z + 4a = 3z + 3a

a = z

In other words, z is 20. That means that when there is NO weight on the spring, it still sits pointing at the "20".

So the actual displacement in the first case (a) is 20 (i.e., the scale "zeroes" at "20", plus another 20 units displacement, to get to 40).

The displacement in the second case (b) is twice that of the first case (it has to be unless the spring fails is does not represent an elastic solid as assumed). The scale zeroes at "20", and the actual displacement is 40, yielding a marking of "60" on the scale.

The spring constant is defined as

F = -kx, where x is the displacement.

100 = -20k (situation 1)

200 = -40k (situation 2)

so k = -5.

Using the unknown weight, the scale reads "30", but we know that the "zero" on the scale is when it reads "20", so the actual displacement is 10. Let U be the known weight.

U = -kx

U= 5 * 10

U = 50 Newtons.

- MadhukarLv 71 decade ago
Increase in weight by 100 N (from 100 N to 200 N) results in increase in reading by 20 (from 40 to 60)

=> for a decrease in reading by 10 from 40 to 30, decrease in weight will be (10/20) * 100 = 50 N (from 100 N to 50 N)

=> X = 50 N.

- Anonymous4 years ago
Sok/mething k/must be neglected of the statek/ment. If the spring is cok/mpressed a distance a and allow move, the field will cok/me to leisure at positions -a and +a with appreciate to the preliminary role, oscillating back/m and forth with angular frequency ?[k/mk/mk/m]. The preliminary deflection a is as you've got calculated.