Question

It takes 6.5 J of work to stretch a spring 4.1 cm from its unstressed (relaxed) length.

a) How much (in J) spring potential energy is then stored by the spring?

b) What is the spring constant k in N/m?

c) What if the spring is compressed instead by 4.1 cm from its unstressed length, how much (in J) spring potential energy is then stored by the spring?

d) How much work (in J) is needed to compress the spring an additional 4.4 cm?

Answer 1

6.5 = 0.5*K*0.041*0.041

K = 7733.49

a) 6.5J

b) K=7733.49

c)
6.5 J

d)

0.5*K( 0.085*0.085 - 0.041*0.041) = 21.43 J extra needed

Answer 2

a) The energy is the same amount of work of external force:

$U = \frac{1}{2}kx^2 = W_{done} = 6.5 J$

b) from equation above:

$k = \frac{2W}{x^2} = \frac{2(6.5)}{(0.041)^2} = 7733.5 N/m$

c) The energy stored is the same if it is compresed or stretch:

$U = \frac{1}{2}kx^2 = W_{done} = 6.5 J$

d) Work needed:

$W = U_f - U_0 = \frac{1}{2}kx_f^2-\frac{1}{2}kx_0^2 = \frac{1}{2}k(x^2_f-x^2_0) = \frac{7733.5}{2}\left ( (0.041+0.044)^2-(0.041)^2 \right )$