Question

A certain spring stores 10.0J of potential energy when it isstretched by 2.00cm from its equilibrium position.How much potential energy would the spring store if it werestretched an additional 2.00cm?How much potential energy would it store if it were compressedby 2.00cm from its equilibrium position?How far from the equilibrium position would you have tostretch the string to store 20.0J of potential energy?What is the force constant of this spring?

Answer 1

Answer:

Answered

Explanation:

x= 0.02 m

E_p= 10.0 J

E_p= 0.5kx^2

10= 0.5k(0.02)^2

solving we get

K= 50.0 N/m

Now

E'_p= 0.5kx'^2

E'_p= 0.5×50×(0.04)^2

E'_p=40 J

b) potential energy is a scalar quantity and it only depends magnitude and not direction so it will remain same in compression and expansion both

c) 20 J = 0.5×50,000×x^2

solving

x= 0.028 m

d) k is 50.0 N/m  from above calculation