Question

A heavy block is suspended from a vertical spring. The elastic potential energy is stored in the spring is 2 J. What is the spring constant if the elongation of

the spring is 10 cm?​

Answer 1

Explanation:

Hey there!

Since mass of the block isn't mentioned i will be considering it to be M.

Consider the initial position i.e. before the block was released to be the dative position (where GPE=0). Now according to the question the block is at rest after displacing 0.1 m.

So using Work Energy Theorem,

W(spring) + W(gravity) = Delta KE = 0  (since the block is at rest both the times)

- 1/2 K (0.1)^2 + M (10) (0.1) =0

K/200 = M

Given the potential energy stored is 2J.

Work Done = Energy Stored

2 = 1/2 * K * (0.1)^2

K = 400 Nm

M = 2Kg

Answer 2

Answer:

A heavy block is suspended from a vertical spring. The elastic potential energy is stored in the spring is 2 J, if the elongation of  the spring is 10 cm the spring constant is $400N/m$

Explanation:

• The elastic potential energy is the energy stored when a force applied to deform an elastic object

• A spring with elastic constant K compressed or expanded at a distance x from its mean position, then the potential energy stored is

        $PE=\frac{1}{2}Kx^{2}$

• The work done by the spring is equal to the change in stored energy

        $W=\frac{1}{2}Kx^{2}$

From the question,

elastic potential energy $PE=2J$

extended length of the spring $x=10cm=0.1m$

the spring constant can be calculated as

$2=\frac{1}{2}K*0.1^{2} \\K=4/0.01=400N/m$