the potential energy of an elastic spring when elongated 12 cm is 3.5 joule. determine the spring constant. what is the potential energy and force of the spring if its elongation is 10 cm?

Answer 1

Let "U" be the spring potential energy. Then:

U = 1/2kx² so if U = 3.5 and x = 12 cm = 0.12 m then:

3.5 = 1/2k(0.12)² or k = 486.111 N/m

So for x = 10 cm (0.10 m):

U = 1/2(486.111)(0.10)² = 2.4 J

F = kx = (486.111)(0.10) = 49 N (in the direction opposite the spring elongation)

Answer 2

Potential energy W = k(Δl)²/2; k = 2W/(Δl)² = 2·3.5J/(012m)² = 486.11 N/m

New potential energy W₁ = k(Δl₁)²/2 = 486.11N/m(0.1m)²/2 = 2.43 J