A certain spring stores11.8J of potential energy when it is stretched by 2.45cm from its equilibrium position
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A certain spring stores11.8J of potential energy when it is stretched by 2.45cm from its equilibrium position

[From: ] [author: ] [Date: 12-06-01] [Hit: ]
8 J (same as extended by 2.23.x = 0.0346 = 3.......
A) How much potential energy would the spring store if it were stretched an additional 2.45cm in J?

B) How much potential energy would it store if it were compressed by 2.45cm from its equilibrium position in J?

C) How far from the equilibrium position would you have to stretch the string to store 23.6J of potential energy in cm?

D) What is the force constant of this spring in N/m?

Please help I would greatly appreciate it!

-
D)
find the spring constant with
E = 1/2*k*s^2
11.8 J= 1/2*k*(0.0245 m)^2
k = 39 317 N/m

A)
E = 1/2*39317*(0.049 m)^2 = 47.2 J

B) 11.8 J (same as extended by 2.45 cm)

C)
23.6 = 1/2*39317*x^2
x = 0.0346 = 3.46 cm


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