Question

If one joule of energy is required to stretch a spring through 20cm, the force constant of that spring is

Answer 1

Answer:

Explanation:

given, streching length(x)=20cm=0.2m and energy(e)=1J

since, e=kx^2/2

          1=k(o.2)^2/2

          0.04k=2

           k=50 N/m

Answer 2

Given:

Energy E = 1J

length x = 20cm

To Find:

Force constant of spring k

Solution:

x = 20cm = 20/100 = 0.2 m

We Know, Energy stored in the spring when it is stretched up to length "x" is given by

E = $\frac{1}{2}$ k$x^{2}$

k = 2E / $x^{2}$

k = 2 * 1 /  $(0.2)^{2}$

k = 2/ $(0.2)^{2}$

k = 2/0.04

k = 50 N/m

So, the force constant of that spring is 50 N/m.