# Two methods for finding spring constant “k”

**1. The problem statement, all variables and given/known data**

A box of mass 5 kg is made to rest on a vertical spring. The box compresses the spring by 10 cm, find the spring constant.

**2. Relevant equations**

I felt this can be solved by two methods :

1. Force of gravity = Outward force of spring

2. The decrease in the potential energy = elastic potential energy stored in the spring.

**3. The attempt at a solution**

The solution in my notes has the first method. So,

** Solution by 1**

F = mg

F = -kΔx … Δx = 10 cm = 0.1 m

=> mg = kΔx

=> (5)(9.8) = k(0.1)

=> k = 490

— — ————–

** Solution by 2**

Initial potential energy of box = mgh

Potential energy at the compressed mg(h-0.1)

Elastic potential energy of the spring = (0.5)k(Δx)^{2}

=> mgh – mgh(h-0.1) = (0.5)k(Δx)^{2}

=> mg(0.1) = (0.5)k(Δx)^{2}

=> (5)(9.8)(0.1) = (0.5)k(0.1)^{2}

=> k = 980

——

So which of these methods is wrong and why?

http://ift.tt/1ct6NMf

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