- #1

Jay_

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**Two methods for finding spring constant "k"**

## Homework Statement

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.

## Homework 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.

## 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

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So which of these methods is wrong and why?

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