Spring constant question: help

[msmitra]

---

Members don't see this ad.

A mass of 1 Kg is placed on a spring and the displacement is 6cm. A mass of 1.5 Kg is placed on the same spring and the displacement is 7cm. What is Hooke's law constant, k, equal to?

A. 2 N/cm
B. 5 N/cm
C. 10 N/cm
D 20 N/cm

To solve this I just set force up (spring force) equal to force down (gravity).

mg = -kX

(1Kg)(10m/s^2) = -k (6cm)

k = 1.6 N/cm

The correct answer is B. 5 N/cm. Why is my method wrong? It seems 100% right to me. I'm confused.

 

[Pisiform]

This is a non-linear spring. If it was linear then its k would be constant like 10/6 = 15/7 but

its not. Nonlinear spring would have a stiffness (k) that varies with x. (unlike ideal, linear

spring that follow Hook's law)

Think of this is like one of those experimental type questions in which you add different

masses to a spring and jot down their extension and then plot the graph of Force vs ext. The

gradient of force extension gives you spring constant (provided LOP is not exceeded). I

wouldn't be a linear graph (because k is different at different points) .. so in reality you

will draw a tangent and find k (not in this question) ...just connect the dots

so co-ordinates you will have (6,10) (7, 15)
Gradient = y2 - y2 / x2 - x1 = 15 - 10/7-6 = 5N/cm

I think this is the way but correct me if I wrong ... I am not good with springs anyway

 

[MT Headed]

This question seems poorly worded. Following the trend, 1.5kg=7cm, 1.0kg=6cm, 0.5kg=5cm, when a 0kg mass is placed on the spring it would "displace" 4cm. Displace from what origin I have no idea.

The X in your "mg=kx" equation is supposed to be the displacement of the spring from its origin. And apparently the spring's origin is at 4cm. So with a 1kg mass, the "mg" side evaluates to 10 newtons, and the "kx" term is now (k)(6cm-4cm).

It seems like this question was written just to be mean, and I never got that feeling from taking AAMC's or the real MCAT.