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3 years ago

A spring stores 10. joules of elastic potential

2 answers:

6 0

The answer would be . Since we are looking for the spring constant you would need to use the formula
$PEs = \frac{1}{2} k {x}^{2}$
. Then you'd substitute, for PEs and x.
$10j=1/2k(.20m^2).$
Then solve. k=500n/m

serg [7]3 years ago

6 0

<h2>Answer:</h2>

The spring constant is 50 N/m.

<h3>Explanation:</h3>

Spring constant is the specific property of a spring which tells about the stiffness of a spring.

It is the amount of force which is used to displace a spring one meter from its equilibrium position.

Its unit is N/m.

The formula of potential energy in a spring is :

P.E = 1/2Kx2

Here,

P.E = 10 joules

x = 0.2m

K = P.E × 2/x2

K = 10 × 2/0.4

k = 50 N/m

Hence the spring constant is 50 N/m.

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